WILEY BENZIGER GEOMETRY UNDER THE EDITORIAL DIRECTION OF ROY DUBISCH AND ISABELLE P. RUCKER LAWRENCE A. RINGENBERG RICHARD S. PRESSER Benziger, Inc. New York, Beverly Hills '',n Association v/itn John Wiley & Sons, Inc. New York, London, Sydney, Toronto EDITORS ROY DUBISCH, Professor of Mathematics at the University of Washington, has been writing for both teachers and students for many years. He has been Associate Editor of the American Mathematical Monthly and Editor of the Mathematics Maga- zine. He has several books to his credit including The Teaching of Mathematics (Wiley, 1963) and is the author of numerous journal articles. Since 1961 Profes- sor Dubisoh has served on the Mathematics Steering Committee of the African Edu- cation Project and has participated in several workshops and institutes in various African countries. He has also directed and lectured at several NSF summer in- stitutes. He is a past Vice President of the National Council of Teachers of Mathematics and has served on the Teacher Training Panel of the Committee on the Undergraduate Program in Mathematics, the Advisory Board and Panel on Supple- mentary Publications of the School Mathematics Study Group, and the Board of Governors of the Mathematical Association of America. He has also been active in many other professional organizations, ISABELLE P. RUCKER, Supervisor of Mathematics of the State Board of Education of Virginia, is a former teacher of both elementary and secondary school mathe- matics. She has participated in summer mathematics institutes and an NSF Academic Year Institute at he University of Virginia and has published numerous articles and book reviews in professional journals. In addition to being active in other professional organizations, Mrs. Rucker has served on the Advisory Board, the Executive Committee, and the Panel on Supplementary Publications of the School Mathematics Study Group, For the National Council of Teachers of Math- ematics, she has served as Chairman of the Committee on Plans and Proposals and as a member of the Committee on Affiliated Groups, She was Program Chair- man for the Golden Jubilee Annual Meeting in 1970 and Is an active member of the Committee on Meetings of the Council. In addition, she has held offices in the Association of State Supervisors of Mathematics, Copyright © 1973 by John Witty and $om, Inc. All nghu r*t»rv*d, PubJfthctf «lmullan»t>im|y )n Canoda. No port of ihl» book may be r»pr«doe»d by any mtoni, nor frflnimi««d, nor troniloted inlo a moctoln* lannyoflfl wifhoul Iho wriltfln p«rmin1on of (h» pvblUhir, ISBN 0-471 -P4549.8 Printed In ihe UnFfsd Sratoi of Afntrka 10 967*34321 AUTHORS RICHARD 5. PRESSER, Coordinator of Secondary School Mathematics of the Michigan City Schools, Michigan City, Indiana, is a former writer for the Minne- sota National Laboratory, the School Mathematics Study Group, and the African Education Program of Educational Services, Inc. Mr, Presser has many years of teaching experience at the secondary school level and is active In the Indiana Council of Teachers of Mathematics and the Central Association of Science and Mathematics Teachers. LAWRENCE A. RINGENBERG, Professor of Mathematics and Dean. College of Letters and Science, Eastern Illinois University, Charleston, Illinois, was Head of the Department of Mathematics from 1947 to J 967. He is a former writer for the School Mathematics Study Group, the author of College Geometry (Wiley, 1968) and Informal Geometry (Wiley, 1967), and is currentry serving as a mem- ber of the editorial board and one of the authors of the forthcoming NCTM Year- book on geometry. Dean Ringenberg was Coordinator of the U.S.A.LD. Mathe- matics and Physics Courses Program in East Pakistan for Intermediate College teachers in East Pakistan during the summer of 1966. Since 1960 he has been a member of the Charleston, Illinois. Unit District Board of Education, Preface Geometry is one of a series of five mathematics textbooks for junior and senior high schools. It is designed for a one-year course for students with a background of informal geometry and elementary algebra such as that included in Mathematics J, Mathematics II, and Algebra of this series. In its formal development of Euclidean geometry, this textbook features an integrated treatment of plane and solid geometry with an early introduction of coordinates. Coordinates on a line, in a plane, and in space relate numbers ami points. They are used to gain knowledge of geometrical figures and to simplify the develop- ment of formal geometry, Students use their knowledge of elementary algebra throughout the book to help them learn geometry. In doing this they maintain and strengthen their competence in algebra. The postulates in this book form the basis of a rigorous, yet plausible, development of a first course in formal Euclidean geometry. In some instances, statements that are proved in more advanced treatments are accepted as postulates here. This has been done to decrease the length of the development and Lo make the develop- ment appropriate for high school students. Geomctrv is an important subject because it is practical and useful and at the same lime abstract and theoretical. There arc two main objectives in this geometry textbook. One is to help students learn a body of important facts about geometrical figures. These facts, interpret* all v. are facts about the space in which we will Preface live. These facts are important for intelligent citizenship and for suc- cess in many careers. The other main objective is to help students attain a degree of mathematical maturity. In the elementary and junior high schools, students are encouraged to learn by observing and manipulating physical objects. There is considerable emphasis On intuitive and inductive reasoning. The power and beauty of mathematics, however, is due primarily to its abstractness. The generality of its theorems makes possible a variety of applications. Understanding mathematics "*in die abstract" is tantamount to understanding the deductive method in mathematics. In studying this book students will develop their capacity to reason deductively and hence their ability to read and write proofs. Contents Chapter 1 Points, Lines, and Planes 1 I.l Informal Geometry 1 / 1.2 The Idea of Formal Geometry 4 / 1.3 The Ideas of Point, line, and Plane 10 / I A Sets 14 / 1.5 Conjunctions and Disjunctions 18 / 1.6 The Incidence Relationships of Points and Lines 22 / 1.7 The Incidence Relationships of Points, T.incs, and Planes 28 / Chapter Summary 34 / Review Exercises 35 Chapter 2 Separation and Related Concepts 40 2.1 Introduction 41 / 2.2 The Betweenness Postulates 41 / 2.3 Using Betweenness to Make Definitions 47 / 2.4 The Concept of an Angle 52 / 2.5 The Separation Postulates 56 / 2.6 Interiors and Exteriors of Angles 67 / 2.7 Triangles and Quadrilaterals 72 / 2.8 Properties of Equality and Number Operations 80 / 2*9 Solving Equations S3 / 2.10 Equivalent Equations 86 / Chapter Summary 92 / Review Exercises 93 Chapter 3 Distance and Coordinate Systems 96 3.1 Introduction 97 / 3.2 Distance 98 / 3.3 Line Coordinate Systems 105 / 3.4 Rays, Segments, and Coordinates 107 / 3.5 Segments and Congruence Contents 111 / 3.6 Two Coordinate Systems on a Line 116 / 3.7 Points of Division 129 / Chapter Summary 135 / Review Exercises 135 Chapter 4 Angles, Ray-Coordinates, and Polygons 138 4.1 Introduction 139 / 4.2 Angle Measure and Congruence 140 / 4.3 Betweenness for Rays 146 / 4.4 Ray-Coordinates and the Protractor Postulate 152 / 4*5 Some Properties of Angles 160 / 4,6 Interiors of Angles 168 / 4.7 Adjacent Angles and Perpendicularity 170 / 4,8 Polygons 177 / 4,9 Dihedral Angles 181 / Chapter Sunn nary 184 / Review Exercises 185 Chapter 5 Congruence of Triangles 188 5.1 Introduction 189 / 5.2 Congruence of Triangles 190 / 5.3 "If-Thcn" Statements and Their Converses 197 / 5.4 The Use of Conditional Statements in Proofs 200 / 5.5 Proofs in Two-Column Form 203 / 5M The Congruence Postulates for Triangles 208 / 5.7 Using the S.A.S., A.S.A.. and S.S.S. Postulates in Wilting Proofs 216 , 5.8 Isosceles Triangles 227 / 5.9 Medians and Perpendicular Bisectors 234 / Chapter Summary 242 / Review Exercises 244 Chapter 6 Inequalities in Triangles 246 6.1 Introduction 247 / 0,2 Inequalities for Numbers 248 / 6.3 The Exterior Angle Theorem 254/6.4 Inequalities Involving Triangles 262 / Chapter Summary 274 / Review^ Exercises 275 Chapter 7 Parallelism 280 7.1 Introduction 281 / 7.2 Definitions 282 / 7.3 Existence of Parallel Lines 284 / 7.4 Transversals and Associated Angles 286 / 7.5 Some Parallel Line Theorems Contents xl / 7.0 The Parallel Postulate and Some Theorems 301 / 7.7 Parallelism for Segments; Parallelograms 308/7.8 Parallelism for Rays 314 / 7.9 Some Theorems on Triangles and Quadrilaterals 318 / Chapter Summary 325 / Review Exercises 325 Chapter 8 Perpendicularity and Parallelism in Space 328 8.1 Introduction 329 / 8.2 A Peipendieularity Definition 332 / &3 A Basic Perpendicularity Theorem ,334 / 8.4 Other Perpendicularity Theorems 333 / 8.5 Parallelism for Lines and Planes 347 / 8.6 Parallelism and Perpendicularity 352 / Chapter Summary 359 / Review Exercises 380 Chapter 9 Area and the Pythagorean Theorem 362 9.1 Introduction 363 / 9,2 Area Ideas 364 / 9.4 Area Postulates 368 / 9.4 Area Formulas 372 / 9.5 Pythagorean Theorem 379 / Chapter Summary 385 / Review Exercises 386 Chapter 10 Similarity 390 10.1 Introduction 391 / 10.2 Proportionality 392 / 10.3 Properties of Proportionalities 396 / J 0.4 Similarities Between Polygons 401 / 10.5 Some I-ength Proportionalities 409 / 10,6 Triangle Similarity Theorems 415 / 10.7 Similarities in Right Triangles 422 / 10.8 Some Right Triangle Theorems 429 / Chapter Summary 436 / Review Exercises 437 Chapter U Coordinates in a Plane 440 11.1 Introduction 441 / 1 1.2 A Coordinate System in a Plane 442 / 1 L3 Graphs in a Plane 447 / 11.4 Distance Formulas 452 / 11.5 The Midpoint Formula 458 / 11.6 Parametric linear Equations 464 / 11.7 Slope 473 / 1 1.8 Other Equations of Lines 481 / 11.9 Proofs Using Coordinates 494 / Chapter Summary 506 / Review Exercises 507 xii Contents Chapter 12 Coordinates in Space 510 12.1 A Coordinate System in Space 511 / 12.2 A Distance Formula 519 / 12.3 Parametric Equations for a line in Space 523 / 12.4 Equations of Planes 526 / 12.5 Symmetric Equations for a Line 539 / Chapter Summary 544 / Review Exercises 544 Chapter 13 Circles and Spheres 546 13.1 Introduction 547 / 13.2 Circles and Spheres: Basic Definitions 547 / 13.3 Tangent Lines 544 / 13.4 Tangent Pknes 567 / 13.5 Circular Arcs, Arc Measure 576 / 13.6 Intercepted Arcs, Inscribed Angles, Angle Measure 584 / 13.7 Segment*: of Chords, Tangents, and Secants 596 / Chapter Summary 607 / Review Exercises 608 Chapter 14 Circumferences and Areas of Circles 612 14.1 Introduction 613 / 14.2 Polygons 614 / 14.3 Regular Polygons and Circles 622 / 14,4 The Circumference of a Circle 634 / 14.5 Areas of Circles; Arc Length; Sector of a Circle 643 / Chapter Summary 636 / Review Exercises 658 Chapter 15 Areas and Volumes of Solids 662 15 .1 Introduction 663 / 15.2 Prisms 664 / 15.3 Pyramids 672 / 15.4 Areas and Volumes of Prisms and Cylinders 680 / 15.5 Volumes of Pyramids and Cones 688 / 15.6 Surface Areas of Spheres and Volumes of Spherical Regions 697 / Chapter Summary 702 / Review Exercises 704 Contents xiii Appendix A-l list of Symbols A- 1 List of Postulates A-2 List of Definitions A-5 List of Theorems A-20 Table of Squares, Cubes, Square Roots, Cube Roots A-42 Glossary G-] Index 1-1 Chapter Sam Folk /Mankmeycr Points, Lines, and Planes 1.1 INFORMAL GEOMETRY Geometry began informally more than 2(KX) years ago in Babylon and Egypt. Tlie meaning of geometry is, literally, "earth measure- ment"; hence it is not surprising thai ruilv knowledge of the subject was concerned largely with measurements of lengths, areas, and vol- umes. Such knowledge was a necessity, especially to the aneient Egyp» tians who almost annually were forced to restore to their river-side farms the boundary markers which were washed away by heavy flood- ing of the River Nile. Then, as well as now, men learned by experience. The ancient Egyptians developed many rules-of-thmnb for surveying fields and roads and for making calculations related to building dwellings and pyramids. These rules were based on many observations, intuition, and reasoning. As long ago as 500 B*G men knew how- to find the areas of rectangles, triangles, and trapezoids, but they had made little progress in formal geometry, that is, in developing geometry as a system that explains why the rules worked. Points, Lines, and Piano* Chapter l As a school subject today geometry is bolh informal and formal In the elementary schools geometry is largely informal. It is physical geometry. Students work with physical objects or with pictures thai represent physical objects, General statement 1 ? or rules are based on intuitive reasoning and inductive reasoning. Intuitive reasoning is what might be called common sense, or, as some might say, reasoning in a hurry. Inductive reasoning is reasoning based on numerous examples. Wc discuss an example of each. Intuitive Reasoning Let us take for granted that we know what is meant by a triangle. Figure 1-! shows a right triangle A ABC with right angle at C figure M Suppose that the lengths of the sides AO and BC are b inches and a inches, respectively. What b the area of the triangle? We say "area of the triangje/' but we really mean the area of the figure or region that consists of the triangle and its interior. In earlier mathematics classes we have learned that the area is \ab square inches. Suppose someone asks why. One answer might be: *"It is intuitively obvious. You can see it by looking at a picture of a rectangle with sides of lengths a and h and with one diagonal drawn." (See Figure 1-2.) If the person who asked why responds with "Now I see why," he is responding to an intuitive feeling of the "tightness" of tilings rather than to a logical argument or to inductive reasoning. This is an example of intuitive reasoning. Figure 1-2 1.1 Informal Geometry Inductive Reasoning Ixl us take for granted that we know what is meant by an angle and its measure. We record the measures of angles as numbers, omit- ting the degree symbol. This is consistent with our formal point of view regarding measure developed later. Suppose that each student in an elementary geometry class is given a set of plastic triangles numl Hired 1, 2, 3, 4, as suggested by Figure 1-3, and a protractor. The instructions direct the students to find and record the measures of the angles of the triangles and then to Hnd the sum of the measures for each triangle as in the following table. The object of the lesson is to make it plausible that the sum of the measures of the angles of a triangle is 180, If the students reason that this is a correct conclusion on the basis of the measurements they have made, this is an example of inductive reasoning. mLA mlB m t C Triangle 1 90 32 58 180 2 75 26 78 179 3 95 25 62 182 4 65 25 90 180 Points, Lines, and Planes : i ,ip-.er 1.2 THE IDEA OF FORMAL GEOMETRY Suppose that as a result of inductive reasoning we htive made a general statement about all triangles, or about all rectangles, or about all circles. We think that the statement is true, but how can we know for certain? Is it possible to know something for certain about all tri- angles? It would seem necessary to test every triangle, which is clearlv impossible. Therefore a different approach is required, the formal approach. The formal approach involves deductive reasoning, that is, logical arguments by which general statements are obtained from previously accepted statements- Two examples follow. Deductive Reasoning Example Suppose that you are given a set of plastic convex quadri- laterals numbered 1, 2, 3 t . . . , 10 as suggested by Figure i-4 Suppose that there are really ten of them, although only four of them are shown. Figure m Suppose also that you know that the sum of the measures of t lie angles of every triangle Is 1 80. Perhaps you know this through the example of inductive reasoning in Section LL At any rate you know or are told this. This knowledge about all triangles is considered as "given" in the situation of this example. The instructions are to find the sum of the measures of the angles of each quadrilateral without making any meas- urements using a protractor. A quadrilateral is convex only if it has the property that if either pair of its opposite vertices is joined by a segment, called a diagonal of the quadrilateral, then all of that diagonal except its endpoints lies inside the quadrilateral. Figure I -5a shows a convex quadrilateral. In Figure 1 -5b the segment PH, except for the points P and R, lies outside the quadrilateral. Hence PQRS is not a convex quadrilateral. A picture of a quadrilateral and one of its diagonals provides a clue for finding the sum of the measures of the angles, (See Figure 1-6.) Quadrilateral ABCD may be considered representative of any convex quadrilateral. From this flgtirc we see that the sum of the measures of 1,2 The Idea of Formal Geometry 5 (a) Convex (b) Not convex Figure 1-5 Figure 1-6 the angles of a quadrilateral is the sum of the measures of the angles of two triangles. Therefore the sum of the measures of the angles of a quadrilateral is 2 * 180 or 360- Without using a protractor we decide that the sum of the measures of the angles of each quadrilateral is 360. We arrive at this conclusion by deductive reasoning. Let us examine this example carefully so that we may understand more clearly the nature of deductive reasoning. We took some things for granted or as given. From these we deduced the answer, or con- clusion. The given statements are called hypotheses, or collectively, the hypothesis. Thus the essence of deductive reasoning is to obtain a conclusion from a hypothesis- The hypothesis may or may not be true. If it is true in a given situation and if the reasoning involved in reaching die conclusion is correct, then the conclusion is true in that situation. Let us look at this idea expressed in symbols. Suppose that wc are given the hypothesis, denoted by If, and we want to establish the con- clusion, denoted by C* What we want to prove is not "C" but "If ff, then C." Many statements in mathematics are in this "If H t then C" form- How can such a statement be useful? It is useful in any situation where the "H " is true. Because if we know "H" is true and if we know "If IT, then C " is true, we also know that "C" is true. In the problem about quadrilaterals, the hypothesis and the conclusion may be identi- fied as follows; H: ABCD is a convex quadrilateral. C: The sum of the measures of the angles of ABCD is 360. 6 Prints, Lines, and Planes If ABCD is a quadrilateral like the one in Figure T -5a, then '*H " is true, and since "If H, then C" is true, it is correct to conclude that "C" is true. If ABCD is a quadrilateral like the one in Figure l-5b, then "H" is false, and although "If ti % then C " is true, it is incorrect to con- clude that "C " is true. Example This example in which deductive reasoning is used to es- tablish a general statement concerns a property of the natural numbers, 1, 2, 3, . . . . Undoubtedly, you know an even number is a number that is 2 times an integer. Thus x is an even number if there is an integer ij such that x = 2y. Some even numbers are 2, 16, 168, and 2466. If you add any two of these numbers, you get an even number. Is it then true that the sum of any two even numbers is an even number? Of course it is. Let us prove it, however, by using deductive reasoning. H: x and y are even numbers. C: x + y is an even number. Our task is to prove: If H, then C. Proof: Since x is even, there is an integer u such that x = 2n. Since y is even, there is an integer c such that y s 2c. Then x + \j = 2u + 2c and, using the distributive property, we get x + y = 2{u + v). But u -+■ 15 is an integer. Therefore, since x -\-yis2 times an integer, x + y is an even number. Notice that we did not prove that x is even or that y is even. We proved that if x and y are oven numbers, then *-f y is an even number. As we said, formal geometry involves deductive reasoning. An im- portant feature of a formal geometry is its structure or arrangement. Hie geometry of this book is elementary Euclidean geometry, carefully arranged so that we can see how the various parts fit together and how some things depend on other things. Formal geometry might be thought of as geometry for the lazy person. In formal geometry we prefer a general statement that tells something significant about all triangles rather than a hundred statements about a hundred triangles. Our starting point in formal geometry is a set of statements about some of the simplest objects of geometry. We do not try to tell what these objects are by definitions. Definitions would involve other words that we would need to define in turn, and so we accept some concepts 1,2 The Idea of Formal Geometry as basic and undefined. We might decide, for example, to start with triangles because almost everyone lias an idea of what a triangle is. But do we actually know what a triangle is? A triangle is made up of three segments. The notion of segment is more basic than the notion of tri- angle. Every segment is a set of points. The idea of a point is more fundamental than the idea of a segment. Every segment is a part of some line. Perhaps a line is simpler to think about than a segment In formal geometry we usually consider points, lines, and planes as the basic building blocks. We do not define these words. How then can we be sure that wc know anytlung at all about them? On the basis of our experience with physical objects we identify the most basic properties that points, lines, and planes have in relation to one an- other. We formulate these as statements that we accept without proof. We call these statements postulates. The foundation for formal geometry, then, is a set of statements, the postulates, which we accept without proof. "The postulates are statements about the basic objects in geometry. We agree that what we know about these objects is what the postulates say, and nothing more, at least at the start. What else can we possibly know about these objects? We can know what %ve formally assume in the definitions and what we deduce by logical reasoning. As an example suppose that we start with the following six postu- lates and one definition. (This is not for keeps— just for this example!) Our notations arc self-explanatory. POST U L ATE A Every line i g a set of p oints and contains at least two distinct points. POSTULATE B Every segment is a subset of a line and contains (besides other points perhaps) exactly two distinct points called its endpoints. POSTULATE C If A and B are any two distinct points, there is exactly one line AB containing A and B, and exactly one segment AB with A and B as endpoints. POSTULATE D Every plane is a set of points and contains at least three points that do not all he on one line, that is, three noncol- linear points. POSTULATE E If A, B, C arc three noncollinear points, then there is exactly one plane containing A, B, and C. POSTULATE F If a plane contains two distinct points A and B, then it contains the segment AB. S Points, Unes, and Planes Chapter 1 Definition If A t B t C arc any three noncolHnear points, then the onion of the three segments AB, BE, and ^A is a triangle; we denote it by A ABC. In this example, what do we know for sure? We know what is said in these seven statements, the six postulates and the definition. What else do we know? We know anything else that we deduce by logical reasoning from them. We shall deduce one statement and call it a theorem. THEOREM Every plane contains at least one triangle. Proof; Let a (read "alpha") be any plane. Then a contains three non- colltiiear points (Postulate D)j call them A, fl, C. Then there is exactly one segment AB with endpoints A and B t exactly one segment BC with endpoints B and Q and exactly one segment Cli with endpoints C and A (Postulate C). Then there is a triangle AABC (by our definition). Each of the segments AB, B£, CA lies m a {Postulate F). Therefore AABC, which is their union, lies in a and the proof is complete. In proving a theorem it is usually a good idea to include one or more inures l.i> suggest the given .situation. In this Mliialirn we start witli 8 plane a as suggested in Figure 1-7, Next we reason deductively to get duee noncollincar points A, B, C in a as suggested in Figure 1-8, Finally, we reason deductively to show that there is a triangle AABC and that it lies in a as suggested in Figure 1-9* Figure 1-7 Our theorem may seem trivial to you. Our goal, however, was sim- ply to show an easy example of a theorem obtained by deductive rea- soning from a set of postulates. 1,2 The Idea of Formal Geomttry EXERCISES 1.2 In Exercises 1-0, a situation is given and a question is asked Obtain an an- swer in each case using intuitive reasoning, 1. Given ihc two numbers x a 21.3 + 27.4 and y = 28.5 + 16.2 + 5.9, which is larger, x or y? 2. Given the two numbers p = (1.6X754) and q = (L6X89G), is pq greater than, equal to, or less than (2)(76SX896)? 3. The following figure shows a square and a parallelogram with side lengths unci angle measures as labeled. Do these figures have equal areas? If not, which one has the larger area? 4 The figure below shows two parallel lines. Is there a plane containing these two lines? «- 5. The figure shows a rectangular box with the vertices labeled. Is there a plane containing the four points ft, G, C* A? 6. Given the situation of Exercise 5, is there a plane containing the four points A, R, C t H? 7. H: C; 8. H: C: 9, H: C: 10. H-. C: TL H: n. 12. H. C: 10 Points, Lines, and Planes Chapter 1 ■ In Exercises 7-12, a hypothesis H and a conclusion C are given. State whether you think the conclusion follows logically from the hypothesis. Be prepared to defend your answer. x is an odd number. x 2 is mi odd number. ar is a multiple of 3. a -2 is a multiple of 6. X is a multiple of 3, x 2 Is a multiple of 9. A ABC is a right triangle with sides of lengths 3 in. 3 4 in. ; and 5 in. The area of &ABC is 6 sq- in. p and q are two distinct intersecting planes. The intersection of p and <] is a line. S is a sphere and p is a plane that intersects 55. The intersection of S and p is a circle or a point. 13 THE IDEAS OF POINT, LINE, AND PLANE Every day in the world around us we observe objects of different sizes and shapes. We notice that some of these objects have corners, edges, and sides, and that some of their parts are "straight," some are "flat," and some are "round." Touching and seeing certain objects help ns to classify them according to their characteristics. Tn arithmetic the idea of a number is a mathematical idea that grew out of a need to classify certain sets according to how many objects they contained. But. no one has ever seen or touched a number. In ge- ometry the ideas of point, line, and plane arc mathematical ideas that grew out of a need to classify certain sets of figures and to measure thei r boundaries or the regions bounded by the figures. But no one has ever seen or touched a point, a line, or a plane. Just as in arithmetic you studied numbers and the operations on them, in geometry you will study points, lines, and planes and how they relate to one another. In the same way that deductive reasoning must be based on certain assumptions (postulates) that we accept without proof so must our definitions be based on certain terms that we make no attempt to de- fine. Tliis is necessary in order to avoid "circular" definitions, that is, a chain of definitions which eventually comes back to the first word being defined. If a person does not know the meaning of any of the words in the chain, the definition is of no value to him. For example, in defining the word "dimension" one dictionary uses the word "mag- nitude." It defines "magnitude" In terms of "size." When looking up 1.3 Th» Idtai of Point, Line, and Plane 1 1 "size" in this same dictionary, it gives "dimension or magnitude," Thus, if someone does not know the meaning of any of the terms mag- nitude, size, or dimension, the dictionary is not very useful. Somewhere in the cycle it is necessary to know the meaning of a word based on experience. The basic undefined terms in our geometry are p>trrf T line, and plane. Our postulates give these terms the meaning that we wish them to have. You undoubtedly already have an intuitive feeling for the con- cept of a point, a line, or a plane. We can think in a vague fashion of a point as having "position" but no "size"; of a line as being "straight;" having "direction," but no "width"; and a plane as being "Hat" but having no "thickness." When we "mark a point" or "draw a line" on our paper or on the chalkboard, we are merely drawing a picture of what we think a point or a line should be, These pictures or figures help us to see and discover some of the relationships that exist among points, lines, and planes and help us to keep these relationships straight in our minds. However, any deductions that we reach must be justified strictly on the basis of our postulates, definitions, and theorems and not on what appears to be true from a figure. As we progress in our study of geometry, we will use figures more and more freely. If our figures are drawn carefully enough, they generally « ill not give us false infor- mation or lead us to false conclusions. Indeed, trying to deduce al the theorems and work all the prob- lems in this text without drawing any figures would be a tedious and difficult task. It would be somewhat like a carpenter attempting to construct a house from memory, without the aid of any blueprints or floor plans. The resulting structure might be quite different from the house he planned to build. We use figures frequently to help explain what our postulates, defi- nitions, and theorems say. You are cneouruged to do the same. When it is practical, you should restate a theorem or a problem in terms of a figure that shows the relationships that arc given in the theorem. If you are careful not to include in the figure any special properties that are not given in the theorem, then the figure should be a valuable aid in understanding the theorem and also in proving it. EXKRCISES 1.3 1. Name three physical objects that convey the idea of a point. 2. Name three physical objects that convey the idea of a line. 3. Name three physical objects that convey the idea of a plane. 12 Points, Lines, and Planes Chapter 1 Exercises 4-9 refer lo the cube shown in Figure 1-10. Answer each question in Exercises 4-7 with the word point, lino, plane, or space, 4. Each cornei' (vertex) of the cube suggests ft!®. 5. Each side (face) of the cube suggests a [7|. 6. Each edge of the cube (such as ATS) suggests 7. The interior of the cube is a subset of [Tj. 8* How many vertices Joes the cube have? How many faces? How many edges? &. Let V represent the number of vertices, E the number of edges, and F the number of laces. Compute V — E + F for the cube. Figure MO In Exercises 10-17, copy and complete the table on the following page for the pyramids and prisms shown below. 14 15. 1,3 The Ideas of Point, Line, and Plane 13 17, i«ZI> E 10. E + F 12. 13. 14, 15. 10. 17. 18. I5id you get the same number for V — £ + Fin Exercises 9-17? If you have counted correctly, V — E ■+■ F equals 2 in each case. Figures like those in Exercises 1CM7 are called polyhedrons. On the basis of your an- swers for Exercises 9-17, do you think that V — E + F = 2 for every polyhedron? Compute V T jE, F, and V — E + Ffor the star-shaped poly- hedron shown in Figure 1-11, (You can construct this polyhedron by gluing together 16 triangles and 2 squares.) Figure MI 14 Points, Lines, and Planes Chapter 1 1.4 SETS In our development of formal geometry we shall often speak of sets of points, or sets of lines, or sets of planes, as well as sets of num- bers. Indeed, we shall agree later in this chapter that every line is a set of points and that every plane is a set of points. Although you are fa- miliar with the language of sets, we s h all review it briefly. Recall that a set is simply a collection of objects. The objects may be numbers, people, points, or whatever. The objects of u set are called the members or the elements of a set. The symbol € indicates that an object is an clement of a given set. Thus, to indicate that the num- ber 2 is an clcmenL of the set {1, 2, 3, 4), we write 2 €{1,2, 3, 4}. Wc read the expression 2 £ {1, 2, 3, 4} as "2 is an element of the set { 1, 2, 3 t 4}" or simply as "2 is in { 1, 2, 3, 4}/' Similarly, if point A is an element of the set of points in line J, we write A£l and read this as "'point A Is on line h" It is often convenient to use set-builder notation to indicate the members of a particular set For example, {x : x is an integer and —2 < x < 4} is just another way of naming the set { — 1, 0, 1, 2, 3}, We read the ex- pression {x : x is an integer and —2 < x < 4} as "the set of all num- bers x such that X is an integer and X k greater than —2 and less than 4." If we wished to include —2 and 4 in this set, we would write — 2 < x < 4 rather than —2 < x < 4. Example 1 Let R represent the set of all real numbers. Suppose thai wc wish to picture (graph) the set C = {x : x € H and -3 < x < 3} on a number line. Another way of writing the expression —3 < x <[ 3 is x ^> — 3 and x < 3. The number line in Figure 1-12 pictures the set A = (s : x € R and x > -3}. ■o— ! i -6 -5 -4 -3-2-1 1 2 3 4 Figure 1-12 1.4 Sets 15 Note that the point corresponding to the number —3 has been circled to indicate that —3 is not an element of this set. The number line in Figure 1-13 pictures the set B = {x : x £ R and x ^ 3}. Why is the point corresponding to the number 3 circled and shaded in the figure? B «- ■S -5-4-3-2-1 I 2 3 4 5 fi Figure 1-1*3 It is clear from the two number lines in Figures 1-12 and 1-13 that the numbers that are common to both sets (those that belong to both A and R) are the real numbers between —3 and 3 and including 3. Tills is shown on a single number line in Figure 1-14. c 4 o ♦ — ! ► -6-5-4-3-2-1 J 2 3 4 5 6 Figure 1-14 We call the set shown in Figure 1-14 the intersection of the two sets in Figure 1-13. The intersection symbol D is used in forming a symbol to denote the intersection of two sets. Thus to indicate the in- tersection of the two sets A = {x : x £ H and x > -3} and B = {x : x £ R and x £ 3} we write A n B and we see that tills intersection is the set C = {x : x £ R and -3 < x <, 3}. rherefore C = A P B. In set language the connective "and" is related to intersection. "Thus the elements that belong to sets A and B in Example 1 are those elements that are common to the two sets; in other words, the inter- section of the two sets is the set C. It may happen that the intersection of two sets is empty, meaimi Li that the two sets have no elements in common. The symbol indicates the empty or null set For example, since the set E of even integers and the set O of odd integers have no integers in common we write En 0=0. In formal geometry, when we say that one set intersects another set, we mean that the two sets have at least one element ra common. 16 Points, Lines, and Planes Chapter 1 In this case, the intersection of the two sets cannot be the empty set. For example, consider the three sets D = {2, 3. 5, 7}, F = {0, 4, 6, 9}, G = (3, 5, 8, 10}. Sets D and F do not intersect since they have no elements in common, and we write DOF= , However, sets D and G do intersect and we write DrC=(3,5). Tn listing the elements of a set, as for D, F, and G, the order in which the elements appear is not important. Thus, if U = {2, 3, 5, 7} } then we may also write D = (2, 5, 7, 3}, The intersection of two sets is in- dependent of order, too. Thus, if A and B are any two sets, then a n n = b n a. Example 2 Let set = {x : x € J* and x < -3 or * > 3}, Let us picture the sets A = {*:x€Rand*< -3} and i?={x:x€fiandx>3} separately on two parallel number lines. (Note: A and B are not the same sets as in Example 1.) The number line (a) in Figure 1-15 shows the graph of the set A = {x : *€ Rand*< -3} and the numt>er line (b) shows the graph of the set 8= {x :x€Randx>3}. A (a) 4 0>) _| i -6 -ft -* -3 -2 -1 i 2 3 4 a u 4- B -► -6 -5 -4 -S -2 -t Figure 1-15 Those numbers that belong to set A or set B are the numbers that he to the left of -3 or to the right of 3. Figure 1-16 on page 17 shows this on a single number line. 1.4 SMs 17 D 4 i . > -6 -5-4-3-2-1 1 a U i 5 6 Figure I- 10 We call the set shown in Figure 1-16 llic union of the two sets in Figiire 1-15. The union symbol U is used in forming a symbol to denote the union of two sets. Thus to indicate 'the union of two sets A = {x : x £ R and x < -3} and B = {% : % € R mix > 3} f we write A U B and see that this union is the set D= {x : *EHand*< -3ori>3). Therefore D = A U B. Example 2 illustrates how the connective "or" is related to unions in the language of sets. Thus the elements that belong to sets A or B are those elements that are cither in set A or in set B or in both. The set of all elements that belong to A Of B is the union of the two sets A and B which, we have seen, is the set D. The union of two sets is independent of their order. Hence, if A and B are any sets, then A J B = B J A, The ideas of union and intersection are extended easily to apply to any number of sets. For example, if A, B, and C are sets, then A n B n C is the set of all elements each of which is in aU three of the sets A. B, and C Describe the intersection of the two sets A and B of Example 2L Example 3 Let A = { l t 2, 3, 4, 5} and B = (3, 4, 5, 6, 7}. The union of A and B is the set of numbers in A or B, and we write AU B= {.1, 2, 3* 4, 5, 6, 7}. It should be noted that if a number is in set A, or in set B, or in both set A and set B, then this number is in A J B. Thus 1 and 2 belong in A U B because 1 and 2 are in set A. The numbers 6 and 7 belong in A U B because 6 and 7 are in set B. Finally, the numbers 3, 4, and 5 belong in A J B because 3 T 4, and 5 are in both set A and set B, The intersection of A and B is the set whose elements are common to both sels A and B, and we write AH B= {3,4.5}. 18 Points, Lines, and Planes Chapter 1 Recall that a set A is a subset of set B if every element of A is also an element of B. (Write A C B for "A is a subset of B") Thus, if A = {«, b), then each of the sets («}, [h], {a, b} t and is a Subset of A. EXERCISES 1.4 L Let A = {x : x£ flandx< lJandB = {x : x£ R andx > -I). (a) Graph the set A. the set B, and the set A (1 B on separate number lines, (h) Use set-builder notation to indicate the intersection of sets A and H. (c) Use set-builder notation to indicate the union of sets A and B. Kxercises 2-8 refer to the sets A= {-3, -1, 1,3},B= (-2.. 0,2, 4}, and C={-2, -1,0,1,2}. 2* Find A n B. A Find A U CL 3. Find A n C. 7. Find A UBUC 4. Find BnC. 8. Find (AtBJfl C. 5. Find A U B, 9. The figure below shows the graph of a set of numbers on a real number hue. Use set-builder notation to describe this set. -7 -6 -5 -4 -3 -2 -1 > 10, The figure below shows the graph of a set of numbers on a real number line. Use set-builder notation to describe this set -"- -6 -5 -4 -3 -a -1 1.5 CONJUNCTIONS AND DISJUNCTIONS Similar to the relationships of two sets and their union and inter- section are the relationships of two statements and their conjunction and disjunction. In mathematics it is very important to understand clearly the relationship of truth in given statements to truth in state- ments formed from the given ones. In this section we consider the rela- tion of two statements to the statements formed from them by con- necting them first by "and" and then by "or." 1.5 Conjunctions and Disjunctions 19 Suppose that p is a statement and that q is a statement. (By a state* ment we mean a sentence that is cither true or false.) We call the state- ment "p and (/"the conjunction of the statement p and the statement q> For example, if p is the statement 4 = 4 and q is the statement 4 = 5 t the conjunction of p and q is the statement 4 = 4 and 4 = 5. We agree to call a conjunction of two statements true if and only if the statements are both true. ► If p is true and q is true, then p and q is true; if p and q is tru e, then p is true and q is true. Thus the conjunction 4 = 4 and 4 = 5 is false because the state- ment 4=5 is false. On the other hand, the conjunction 4=1 unci 5 = 5 is true. Of course, if both of the statements p and q are false, then the conjunction p and q is false. Let us see how the conjunction of two statements is related to the intersection of two sets. In Example 1 of Section 1.4, we considered the intersection of two sets A = {« i x £ R and x > -3} and B = {x : x £ R and x < 3}. We saw that the intersection of these two sets is the set of aE numbers that belong to both A and B. In other words, if X is a number, then it is true that x is an element of A and of 8 (x £ A D B) if and only if it is true that x is an element of A and it is true that x is an element of B. For example, 2 £ A and 2 £ B; hence 2 £ A P B. On the other hand, 5 £ A and 5 £ B (5 is not an element of B), hence 5 £ A n J3. Thus if the statements i£A and x £ B are true, then the conjunc- tion of these statements (x £ A n B) is also true. But if either of the statements x £A and * £ B is false, then * € A n J? is false. Now consider the statement p or q where p and q are statements. We call the statement p or q the disjunction of the two statements p and C/. For example, if £? is the statement 4 = 4 and f/ is the statement 4 = 5, the disjunction of p and q is the statement 4 = 4 or 4 = 5. 20 Points. Lines, and Planes Chapter I We agree to call the disjunction of two statements true if and only if either, or both, of Ihc two statements is true, ► If /> is true and q is true, then p or q is true; if p is true and q is false, then povq is true; if /> is false and q is true, then p or q is true; if p or q is true, then either p is true and q is false, or p is false and q is true, or p and q are both true. Thus the disjunction 4 = 4 or 4 = 5 is considered to be true be- cause 4 = 4 is true. Similarly, the disfunction 4 as 4 or 5 s 5 is t me since the statements 4 = 4 and 5 = 5 are both true. On the other hand, the disjunction 4 = 5 or 7 < 5 is False since I Kith of the state- ments 4 = 5 and 7 < 5 are false. Let us see how the disjunction of two statements is related to the union of two sets. In Example 2 of Section 1.4 we considered the union of the two sets A = (x : x € B and x < -3} and B = {x : x £ R and * > 3}. We saw that the union of these two sets is the set of numbers belong- ing to either A or B. In other wortfc. if x is a number, then it is true that i is an element of A or B (x € A U B) if and only if at least one of the statements x is an element of A and x is an element of B is true. For example, 5 € A, but 5 € B; hence 5 C A U R On the other hand, 2 g A and 2 £ B; hence 2 t A U B, Thus, if at least one of the statements x £ A and x £ B is true, then the disjunction of these two statements (x £ A U B) is also true. But if both of the statements x € A and x £ B are false, then the statement x 6 A U B is false. FAERCISES 1.5 Exercises 1-6 refer to Figure J -17. Imagine that each of the four lines a, b, c.din the figure is a set of points. FJeuru I- 17 3 .5 Conjunctions and Disjunctions 21 L The set {£} is the intersection of which two lines? 2. What is the intersection of h and c? 3* What is the Intersection of the three sets a t h, and rf? 4 Does « fl fc = a P fi? 5, What is a H c? 6. What is the intersection of the four sets a, b, c, and d? The following table shows the possible truth values (T for true, F for false) that can be assigned to the statements p and q. In Exercises 7-10, copy and complete the table for the statements p and q and p or q. p and q p or q 7. T T T 8. T F F ft. F T CD 10. F F DQ 11. If x £ A and x £ B t is ran element of A H B? Of A U B? In Exercises 12-15. P and Q represent sets. Copy and complete the table using T for true or F for false. *€ P x€Q x 6 P n Q XC p \J Q 12. T T T 7] 13. T F a m 14. F T 3 3 15. F F 13 F 16. Two sets arc said Lo be equal if they contain exactly the same elements. If A n {x ; ar is an integer and - 1 < % < 5} and B - {0, 1, 2 S 3, 4), does A = B? Why or why not? If C = {2. 3, 0, 1 , 4 } } does R = C? 17, In Exercise 1.6 is A a subset of B? Is B u subset of A? Is 5 a subset of C? Is C a subset of B? IS. If A C B and B C A are both true, what can you conclude? 19. If A and B arc sets and A = B i does B = A? 20. If A and B arc sets and A C B, is B C A? 22 Points. Lines, and Planet Chapter 1 21. (a) List all the subsets of the set { 1, 2, 3}. (Remember that the empty set is a subset of every set.) (b) How many distinct subsets does a set co nsisting of three elements have? Four elements;? n elements? 22. If A = and & = {3, 4, 5, 6}, what is A H B? A J F? 23. (a) Given that x is a number such that x 2 = 16, find one possible value for x. Is there another possible value for x? What is the solution set of the equation x 2 = 16? (b) Write die solution set of the equation x 2 + 7 = 56. ■ In Exercises 24-29, write the solution set of the equation. 24. 3a - 5 = 13 27. 2(* + 3) = 5a - 8 25. 5(2* - 9) = 10 28. x* - 5 = 31 26. 3(2* - 7) = -t + 9 29. x 2 - 2x - 15 = 30. If F = f (x, y) : x £ H, tj C R. and y = 2x - 3}, we find that when i=2 T y = 2'2 — 3 = 1. Thus the ordered pair (2, 1} is an element of the set F. To check this, note that "2 £ R, 1 € R, and 1 = 2 ■ 2 - 3" is a true statement. This is the statement you get if you replace x by 2 and y by 1 m the sentence which follows the colon in the set-builder symbol (a) What value of tj is paired with x when x = ^? (b) Find five more ordered pairs of numbers that belong to the set F. 31. Describe the solution set of the equation 2(x — 2) = 2x — 4. 32. Describe the solution set of the equation 2{x — 2) = 2x - 5. 1.6 THE INCIDENCE RELATIONSHIPS OF POINTS AND LINES We are now ready to begin the formal development of geometry. We do not state all the postulates of our formal geometry at once, but rather introduce them throughout the text as needed. In this chapter. we state the first eight of que postulates. These postulates are con- cerned with points on lines, in planes, and in space; lines through points, in planes, in space ; and planes through points, through lines, and in space. Figure 1-18 shows a picture of a line I and a point F. As the figure suggests, point F is on line / and line / passes through point F. Since t is a set of points and P is an element of this set, it is correct to say that F is "in" I or that F is a point "of" L However, we usually say that Figure 1- 1,6 Relationships of Points and Lirv« 23 Pis "on" I and that 1 "passes through" P, Another way of saying this is to say that the point and line are incident, it is for this reason that our first postulates, which we state in this section and in Section 1.7, are called Incidence Postulates: For our first postulates we draw on our experience with physical representations of points and lines. When you use your ruler or straightedge to draw a line through two points, such as P and Q in Fig- ure 1-19, you are, of course, actually working with physical representa- tions of points and lines and not the abstract points and lines of our geometry. Figure 1-19 We decide what we are going to accept without proof about geo- metric points and lines by looking at the real world. In fact, we need only look at Figure 1-19 to arrive at our first three postulates. First, however, wc state two definitions. Definition 1.1 Space is the set of all points Definition 1.2 The points of a set are collinear if and only if there is a line which contains all of them. The points of a set are n uncoil in ear if and only if there is no line which con- tains all of them. Plane Postulates of Incidence POSTULATE 1 (The Tiiree-Point Postulate) Space contains at least throe noncollinear points. POSTULATE 2 (The Line-Point Postulate) Every line is a set of points and contains at least two distinct points. POSTULATE 3 (The Point-Line Postulate) For every two distinct points, there is one and only one line mat contains both points. 24 Points. Lin«. and Planes Chapter 1 Postulate 3 is sometimes shortened to read "two points determine a line." When we use the word "determine" here, we use it in the sense of Postulate 3, which states that, given two distinct points, there is exactly one line which contains them. Similarly, when we say that "three noncolliiicar points determine three lines," we mean that there are exacdy three lines each containing two of the three points. You may feel that these postulates do not tell us very much about points and lines. Your experience has led you to believe that there arc a great many points on a line and in space, and you may wonder why we do not say so in the postulates. We feel that it is more instructive to proceed as we have done, Shortly we will have introduced enough postulates that we will be able to prove that there are infinitely many points on a line and in space. However, there is not much we can prove from only the three postulates. Figure 1-19 should suggest to you at least two statements concerning lines that are not stilted in the postu- lates. Before formulating these statements, we will introduce some Symbols for points, lines, and planes. Notation. Capital (upper case) letters are used to denote points and small (lower case) letters to denote lines. For example, in Figure 1 -19 P, Q, and R denote points and /, m, and n denote lines. We may also name a line by naming two points that determine the line. Thus cither of the symbols PQ or Qr could be used to name the line that is deter- mined by points P and Q. It may also be convenient to name several different lines by using the same letter with distinguishing subscripts such as h, I2, fa* * • • « In the figures we represent a line by the symbol 4 ► where the arrowheads are used to indicate that the line does not stop where our picture stops but continues indefinitely in both directions. We often use the Creek letters a, /?, v, . , , (alpha, beta, gamma, . . .) to name a plane, Sometimes it is convenient to name a plane by nam- ing three noncollinear points that the plane contains, such as "plane FQfi* in Figure 1-19. Let us return now to our discussion of the relationships between points and lines. Figure 1-19 suggests, but our postulates do not tell us, that there are at least three distinct lines in space. Let us now at- tempt to deduce this statement from the three postulates we already have. The Three- Point Postulate tells us that there are at least three points in space that do not all lie on one line. For convenience we name these points P t Q> and R. It follows from the Point-Line Postulate that 1.6 Relationships of Fbints and Lines 25 points P and Q determine exactly one line PQ, that points Q and R de- terminc exactly one line QR, and that points P and R determine exacdy one line PR. No two of these lines arc the same. For if they were, points P, Q, and R would be collinear, and this would contradict the fact that P, {), R do not all lie on one line. This completes the proof of the fol- lowing theorem. (Usually the proof of a theorem follows the statement of the theorem. For Theorem 1.1, however, the proof precedes the statement of the theorem.) THEOREM 1, 1 There are at least three distinct lines in space. We observe from Figure 1-19 that no two distinct lines intersect in more than one point We proceed now to deduce this statement from our postulates. Tf /, and t 2 are any two distinct lines that intersect, then, by defini- tion of intersect, we know they have at least one point in common. Call this point P. Now, either they have a second point in common or they do noL Let us suppose that they do have a second point Q in com- mon as suggested In Figure 1-20. Figure 1-20 The Point-Line Postulate tells us that for every two distinct points there is one and only one line that contains them. Thus h = h (diat is, they are the same line). Rut this contradicts our hypothesis that k and h are distinct tines. Since our supposition thai l 1 and fe intersect in a second point leads us to a contradiction, we must conclude that h and I2 intersect in no more than one point. We have proved the following theorem. THEOREM 1,2 If two distinct lines intersect, then they intersect in exactly one point. Does Figure 1-20 confuse you because the marks that represent lines do not look as you think straight lines should look? Actually the word straight is not part of our formal geometry. We want a formal 26 Points. Lines, and Plans* Chapter 1 geometry of points, lines, and planes that is consistent with our experi- ences in informal geometry. In particular, we want lines in our formal geometry to have the property of straight ness. It is true that we cannot draw a convincing picture of two distinct lines with two distinct points in common. Tills is evidence at the informal geometry level consistent with Theorem 1.2. EXERCISES 1.8 L Winch postulate asserts that there are at least two distinct points on everj' line? 2. Restate the Line-Point Postulate and the Point-Line Postulate in your own words. 3. Restate the definition of "intersect" as the word applies to two sets of points. 4* If points A, B, C, and D are distinct, and if line / contains points A, B, and C and line in contains points A, C, and D t what can you conclude from this? 5. In proving Theorem 1 . 1 wc started with three noncollincar points and showed that there are three distinct lines each containing two of the three noncollincar points. Suppose we start with four points, no three of them collincar. How many distinct lines are there each containing two of the four points? 6. Suppose we .start with five distinct points, no three of them coBmear. How many distinct lines are there each containing two of the five points? 7. Suppose we start with six distinct points, no three of them collincar. How many distinct lines are there each containing two of the six points? 8. Consider your answers to Exercises 5, 6*, and 7. Can you predict how many lines there are for seven distinct points each containing two of the seven points if no three of the points are collinear? If so, how many? 9. Let m and n lie different lines. Let A be a point snch that A g m and A <E n, Let B he a point such tliat B £ m and B C «• What must be true of points A and B? What postulate or theorem supports your answer? 10. Let A and B be different points. Let line i% contain A and B and let line /a contain A and B, What can you conclude iihout / t and l z ? What pos- tulate or theorem supports your conclusion? 1.6 R#l»tiofl ships of Points and Lines 27 11. Consider the set consisting of the five points,, and no others, named in the following figure. If the points appear to be eollinear, assume that they are, (a) Identify, by listing the members, two subsets each containing three eollinear points. (b) Identify, by listing the members, four subsets each containing four noneollmear paints of which three points are unilinear. (c) Ust the members of one subset contai ning four noneollinear points of which no three are eollinear. (d) Name three lines which are not drawn in the figure, but each of which contains two points named in the sketch. 12. From which postulate does it follow that no line contains all points of space? challenge pnonLUM* Start with three distinct objects such as three boys named Jerry, Jim, and John, Create a structure as follows: Each boy is a point and there is no other point. Thus space = {Jerry, Jim, John}. There are three distinct lines a, h, and c (and no others) as follows: a = {jerry, Jim}, b = {Jim. John), c = (Jerry, John}. Exercises 13-30 are questions about the Jerry- Jim-John structure. Kxercises 31-34 are general questions. 13* Are the points, Jerry, Jim, and John, eollinear? 14. Does space contain at least three noncollinear points? 15. Does the structure satisfy the Three-Point Postulate? 16. Is a a set of points? 17. Does a contain at least two distinct points? 18. Is b a set of points? 19. Does h contain at least two distinct points? 20. Is c a set of points? 21. Docs c contain at least two distinct points? 22. Does the structure satisfy the Line- Point Postulate? 23. Is {Jerry, J tin} a set of two distinct points? 28 Points, Lines, and Planes Chapter 1 24, Is {Jerry, John) a set of two distinct points? 25. Is {Jim, John} a set of two distinct points? 26. Are there other sets of two distinct points? 27, Is there one and only one line that contains Jim and John? 2S, Is there one and only one line that contains John and Jerry? 29, Is there one and only one line that contains Jerry and Jim? JO. Does the structure satisfy the Point- Line Postulate? 31, Can we prove, using only the plane postulates of incidence, that space contains at least four distinct points? 32, What do the Plane Postulates of Incidence tell us about the nature of an individual point? 33, What do the first three postulates tell ns about the straightness of a line? 34, What do the first three postulates tell us about segments? 1.7 THE INCIDENCE RELATIONSHIPS OF POINTS, LINES, AND PLANES We are now ready to list our remaining Incidence Postulates. We begin by stating the following useful definition. Definition 1.3 The points of a set arc eoplanar if and only if there is a plane %vhich contains all of them. The points of a set are noncxmluiuir if and only if there is no plane which con- tains all of them. Space Postulates of Incidence POSTULATE 4 (The Four-Point Postulate) Space contains at least four nonco planar points. POSTULATE 5 (The Plane-Point Postulate) Every plane is a set of points and contains at least three noncollinear points. POSTULATE C (The Point-Plane Postulate) For every set of three noncollinear points there is one and only one plane that concur them, POSTULATE 7 (The Flat-Plane Postulate) If two distinct points of a line belong to a plane, then every point of the line belongs to that plane. POSTULATE 8 (The Plane-Intersection Postulate) If two dis- tinct planes intersect, then their intersection Is a line. 1,7 Relationships of Points, Lines, and Planes 29 Note that our postulates do not include a statement about the num- ber of planes in space. Theorem 1.3 states that there are at least two distinct planes in space. In the Exercises that follow you will be asked to prove that there are at least four distinct planes in space. From Postulate J we know that there arc aL least three noneollincar points in space. Call these points A, B, and C, By Postulate 6 there is exactly one plane a that contains these points. From Postulate 4 there is a fourth point D that is not in plane a. Points A, B t and D are not unilinear, for if they were, point D would lie in plane a. Why? By Postulate 6 again there is exactly one plane /? that contains points A, B, and D. Planes a and f$ are not the same plane, for if they were, points A, B, C. and D would be coplanar, which is contrary to the way they were chosen. We have proved the following theorem. THEORF.M L3 Space contains at least two distinct planes. It is often helpful to draw diagrams or pictures showing the rela- tionships between points, lines, and planes when making; a deduction such as the one for Theorem 1,3. Figure 1-21 shows points A, B, and C in plane a and points A, B. and D in plane /?. (Although we often use a quadrilateral to represent a plane, be careful not to think of the sides of the quadrilateral as the "edges" or "ends" of the plane.) Figure 1-2] also helps us to "see" that when two planes intersect, their intersection is a line as we have stated in Postulate 8. What is the intersection of a line and a plane not containing the line? When we look at Figure 1-22, it appears that the intersection is a single point L Figure 1-21 1 Figure 14S 30 Points, Linti, and Planet Chapter 1 THEOREM 1.4 if a line intersects a plane thai does not contain the line, then the intersection is a single point. We can prove Theorem 1 ,4 by the same technique used in proving Theorem 1.2. Head again the proof of Theorem 1.2 which appears hefore the statement of the theorem. Now try to answer the questions in the argument (proof) that follows* By hypothesis and the definition of intersect, line m and plane of (Figure 1-22} have at least one point F in common. How do we know that plane a does not contain every point of line m? Suppose line mi and plane a have a second point Q in common. (Here we are examining one of only two possibilities: either the line and plane have a second point in common or they do not. Of course, we are trying to prove that they do not.) But if a plane contains two distinct points of a line, then it contains the entire Line. (What postulate are we using here?) What does this last conclusion, namely, that the plane contains the entire line, contradict? Since we have agreed to accept our postulates as true statements al>out points, lines, and planes, what must we con- clude? Docs this prove the theorem? It follows from Postulate 6 (the Point-Plane Postulate) that three noncoUinear points determine exactly one plane. That is, there is exactly one plane which contains any set of three noncoUinear points. We conclude this section with two additional theorems regarding the existence of planes. THEOREM 1.5 If m is a line and P is a point not on m, then there is exactly one plane that contains m and P. Note that we must prove two tilings. 1. There is at least one plane t hat contains m and P. & There is no more than one plane that contains m and P. These two statements illustrate the ideas of existence and uniqueness. When we prove existence, we show that there is at least one plane con- taining m and P. When we prove uniqueness, we show that there is at most one such plane. If we can prove both existence and unique- ness, then we know that there is exactly one such plane. In the following proof name the postulate or theorem that justifies the statement preceding the question "Why?". Proof of I. Existence. There is a plane a that contains line m and point P (Figure 1-23). 1.7 Relationships of Points, lines, and Planes 3 1 There arc at least two distinct points on line m. Why? Call these points A and B. The points A and B do not lie on any line except m. Why? By hypothesis, point P does not lie on line m. Hence points A, B, and V arc noncoUinear. There is a plane tt Lhat contains points A, B t and P. Why? If plane a contains points A and B, then it contains line ro. Why? Hence there is a plane that contains line m and point P, Mgurc 1-23 Proof of 2. Uniqueness. There is no more than one plane tJiii contains line m and point P. Suppose that there is a second plane /? which contains m and P. Then /? contains the noncoUinear points A, B t and P. Thus we have two distinct planes* a and j8 s each containing three noncoUinear points. What postulate does this contradict? Does litis prove that there is no more than one plane containing m and F? THEOREM L6 If two distinct lines intersect, then there is exactly one plane that contains them. Let lines m and n intersect in the point P (Figure 1-2-4). a Figure 1-24 We must prove the following: 1. Etafatencft There is at least one plane that contains Hi and m. 2. Uniqueness. There is no more than one plane that contains m- and it, 32 Points, Lints, and Planes Chapter 1 In the following proofs name the postulate or theorem that justifies the statement preceding the question "Why?" Proof of 1; Lines m and n have exactly one point in common. Why? There is a point A on m that is different from P t Why? A is not on n by Theorem 1.2. (Two distinct lines intersect in no more than one point.) There is a plane a which contains A and n. Why? Plane a contains F and A and therefore contains m. Why? Hence there is a plane a which contains m and n. Proof of 2; Suppose there is a second plane that contains both m and n. Then plane contains A because A is on m. Thus both a and fi contain A and n. What theorem docs this last statement contradict? Does this prove that there is no more than one plane a containing m and n? EXERCISES 1.7 Refer to Figure 1-25 in working Exercise* 1-12. Assume that points A t B, and C are in the plane of the paper and that point D is not Figure 1-19 1* Name all the planes, such as ABC, that are determined by the points in the figure 2. Name atl the lines that are determined by the points in the figure. One of them is line A£. 3. How many lines arc there in each plane? 4. How many planes contain line BD? Name them. Is this number of planes the same for every line? 5. How many lines are on every point? Name the lines that are on point D. 6. How many points do plane ABC and line 55 have in common? What theorem justifies your conclusion? 1.7 Relationships of Points, Lines, and Planes 33 7. Name two planes, unci name a point that is contained in both planes. Which postulate tells ils that these two planes have a line in common? Name the line. 8. There are at least how many points common to every pair of planes in the figure? 9- Two planes which do not intersect arc said to be parallel. On the basis of our first eight postulates, do you think it can be proved that there must be two parallel planes in space? 10, Is it possible for three planes to have one and only one point in com- mon? If your answer is "Yes/' name three planes and the single point they have in common. 11, Two noncoplanar lines in space are said to be skew lines. Name three pairs of lines in the figure such that each pair is a pair of skew lines. Can a pair of skew lines intersect? Why? 12. Can three lines have a single point in common and be noncoplanar lines? 13, Let P be a point in plane a. On the basis of only our first eight postu- lates do you think that we can conclude that there are three distinct Hues in plane a that contain point F? Give reasons for your answer. 14. How many points do the planes of the ceiling, front wall, and side wall of your classroom have in common? On the basis of only our first eight postulates do you think it can be proved tint there arc three distinct planes that have more than one point in common? 15. Using three pages of your book as physical examples of planes, what would you guess to be the intersection of three distinct planes if then- intersection contained more tlian one point? 16\ I Ay the edge of your ruler along the top of your desk. Do all the points on the edge of the ruler seem to touch the desk? Which of our postulates docs this illustrate? In Exercises 17-25, write a short paragraph to show how our postulates or theorems can be used to prove the .statement. 17. Every line is contained in at least one piano. 18. There are at least three distinct lines in every plane. 19. There are at least four distinct planes in space. 20. There are at least six distinct lines in space. 21. If four points are noncoplanar, then they are noneollinear. 22. If four points are noncoplanar, then any three of these points are noneollinear. 23. There are at least three distinct lines through every point. 24. Every line is contained in at least two distinct planes. 25. Every point is contained in at least three distinct planes. 34 Points, Ones, and Planes Chapter i 26. challenge problem. Suppose, for this one exercise, that we replace our Postulate 8 with this postulate: "If two distinct planes have one point in common, then they have a second point in common." Prove our replaced Postulate 8 as a theorem. 27. challenge problem. Suppose, just lor Exercises 27 and 28, that space consists of four distinct points A, B, C, D and no others. Suppose that there is one line a containing all four of the points A, B, C, D and that there are no other linos. Suppose that there are no planes. Does this structure satisfy Postulate 4? Since A, B, C, and D are not contained m any plane, it follows that they arc noncoplanar, and hence Postulate 4 is satisfied. Does this structure satisfy Postulate 5? Yes, it does. To show lliis we must check each plane in the structure to see if it contains three noncollinear points. If there is no plane that violates the property of Postulate 5, then Postulate 5 is satisfied. Since there is no plane in the structure, it follows that there is no plane that violates Postulate 5, and hence Postulate 5 is satisfied. Show that Postulates 6, 7, and 8 arc satis- fied by this structure, 28. challenge p itonLEM . Does the structure of Exercise 2 7 satisfy Postu- late 1? Postulate 2? Postulate 3? Your intuition might tell you that Pos- tulate 4 includes Postulate 1 in the sense that Postulate 1 could he proved once Posulate 4 is assumed. The structure of Exercise 27 pro- vides an answer to the question; Is it possible to satisfy Postulate 4, in- deed all of the postulate, from 2 through 8, without satisfying Postulate 1? In other words, is Postulate 1 independent of Postulates 2 through 8? Why? CHAPTER SUMMARY In tliis first chapter we compared informal geometry and formal ge- ometry noting that geometry began informally many years ago as a collec- tion of rules to solve practical problems related to "earth measurement." Although geometry as a school subject is both formal and informal, the ap- proach in tins book is predominantly formal. Tile formal approach features carefully stated postulates, definitions, and theorems. Proving theorems involves- deductive reasoning. We dis- cussed examples of intuitive reasoning, inductive reasoning, and deductive reasoning. We reviewed the language of sets. If A and B are sets ? the INTERSEC- TION of A and B, denoted by A n B t is the set of elements contained both in A and in B, The UNION of A and £, denoted by A U B, is the set of ele- ments contained in A or B, The intersection of two sets may be die EM PTY or NULL SET denoted by . If two sets INTERSECT, they must have at least one element in common. Review Etercis** 35 If p and q arc statements; the CONJUNCTION of these two state- ments is the statement p and q. The conjunction of p and o; is true if both p and ^ arc true, but false in all other cases. The DISJUNCTION of the two statements is the statement porq. The disjunction of p and q is false if both p and q are false, but true in all other cases* In order to avoid circular definitions w« accepted POINT, LINE, and PLANE as UNDEFINED TERMS in our formal geometry. We defined SPACE to be the set of all points. The points of a set are COLLINEAR if and only if there is a line that contains all of them. The points of a set are COPLANAR if and only if there is a plane that contains all of them. Our first eight postulates are called INCIDENCE postulates and arc listed below by name only. Be sure that you can state each of them in your own words. 1. THE THREE-POINT POSTULATE 2. THE LINK-POINT POSTULATE 3. THE POINT-LINE POSTULATE 4. THE FOUR-POINT POSTULATE 5. THE PLANE-POINT POSTULATE 6. THE POINT-PLANE POSTULATE 7. THE FLAT-PLANE POSTULATE 8. THE PLAxNE-INTEftSECnON POSTULATE In addition to these eight postulates we stated and proved six theorems. Read these theorems again and study their proofs. REVIEW EXERCISES In Section L I an example involving the area of a right triangle was given to illustrate intuition or iutuitive reasoning. In Exercises 1-10, state whether or not you think the given example is a good example of intuitive reasoning. If you think the decision made is false, indicate it. 1. Examine a rectangular box and decide that the opposite faces of a rec- tangular solid have the same size and .shape. 2. Examine a ball and decide that a plane through the center of a spherical region (a Sphere and its interior) would separate it into two hemispheri- cal regions with equal volumes. 3. Examine a ball and decide that the area of a sphere is four times the area of a circular cross section formed by a plane passing through the center of the sphere. 4. Examine three distinct points on a line and decide that if A, B y C are any three distinct points of aline with B between A and C, then the distance from A to H is less than the distance from A to C. 36 Points, Unas, and Pfants Chapter 1 5. Think about a triangle Trying in a plane a. Think about a line in or that passes through a point F lying inside of T. Decide that this line must contain two distinct points of T. 6*. Think about a triangle, a quadri- lateral, and a pentagon. Decide that the sum of the measures of the angles is the same for each. 7. Think about three books of 300 pages each standing upright on a shelf as suggested in the figure. De- cide that it is farther from page 1 of book I to page 300 of hook 1 II than it is from page 300 of book 1 to page 1 of book III. 8, Think about a plane figure consisting of a triangle and the bisector of one of its angles as suggested in the following figure. Decide that the bisector of an angle of a triangle divides the opposite side into two seg- ments of equal length. 9. Tliink about a circle and a line that separates it into two arcs of equal length. Decide that the line passes through the center of the circle, 10. Draw a triangle Mark one point on each side as close to the midpoint as you can without measuring. These points are vertices of a second triangle. Decide thai the area of the original triangle is four times the area of the second triangle. In Section LI an example involving the sum of the measures of the angles of a triangle was given to Illustrate inductive reasoning. In Exercises 11-20 draw one or more figures and make some measurements if you wish. Then draw an appropriate conclusion. 11. Is the following statement true or false? The bisectors of the angles of a triangle are concurrent, thai is, they all pass through the same point. 12. What is the sum of the measures of tlie angles of a pentagon? (Consider only pentagons whose diagonals pass through the interior of the pentagons.) 13. If A BCD is a quadrilateral with AB = BC, h it necessarily true that CD = DA? Review Exercises 37 14. If ABCD is a quadrilateral with AC = HD t is it necessarily true that AB = BC? 15. If the sides of a quadrilateral have equal lengths, is it necessarily true that its diagonals have equal lengths? 16. If the diagonals of a quadrilateral bisect each other, is the quadrilateral necessarily a parallelogram? 17. If the sides of AABC have equal lengths, if D is the midpoint of BC, If E is on line DA, if D is between A and E» and if AD — DE, is it true that S?is parallel to AC? 18. If AABC is any triangle and if D is the midpoint of EC* is it necessarily true that AABD and A A CD have equal areas? 19. If AABC is a triangle and if D is the midpoint of B~C, is it true that AD = %AB + AQ? 20. If A ABC" is a scalene triangle (no two of its sides have the same length), is it true that no two of its angles have equal measures? Section 1,2 contains two examples of deductive reasoning. In Exercises 21- 25, state whether you think the conclusion C (Ci, Ca, . . - , if more than one) follows logically from the hypothesis H (Hi, 91% if more than one). Be prepared to explain what assumptions you are making, if any, in addition to those that are stated as hypotheses, 21. H. x = o C: 2* -f- 3 = 13 22. Hi 2x + 3 = 13 Ct x = 5 23. Hi'. Every polygon is a plane figure, ll'i- P is a polygon. & P is a plane figure. 24. B Let AABC and ADEF be given. If ml A = mLD, mlB = B»Z£, and AB = DE, then AC = DFsad BC = EF. G; tx;t AABD be given with C between A and D. If m / ABC = ml VBC and mlACB = m I DCB, then AC = CD Q: Let AABC and ADEF be given. If ml A ~ mZD, AB = DE f and AC = DE, then BC = EF. 25. Hi: ATI cows eat grass. H& Bessy is not a cow. Ct Bessy does not eat grass, 26. If A is the set of even integers and B is the set of odd integers, describe the union of A and B< the intersection of A and B. Do sets A and B Intersect? 38 Points, Lines, and Plane* Chapter 1 ■ Exercises 27-38 refer to the three sets Q= {-3, 0, 3 r G), <J= {-3, 0, 3, 6,9}, T = [x \ x is an Integer, - 3 <; x < 9, and x is divisible by 3), 27. Is Q C S? 33. Is T C S? 28. Is S C 9? 34. Docs S = 7? 29. Is Q c 7? 35. Find Q D S. 30. Is T C 0? 3fl - Uoe* D S = Q? 31. Does Q = T? 37. Find Q U S, 32. Is S c T? 38. Does £ U S = S? 39. If A = {x ; x £ R and x < 7} and B = (x : x £ /I and x > 3}, draw the graphs of A, B, and A H Bon three separate number lines. 40. Describe the set A H /J in Exercise 39 using set-builder notation. 41. If S = {x : x £ R and x > 5} and T = {x : x C R and x < 0), draw the graphs of S, T, and S U Ton three separate number lines. 42. Describe the set S U T in Exercise 41 using set-builder notation. ■ If p is the statement 7 = 7, q is the statement \/9 as 5, and r is the statement — 5 < —2, determine in Exercises 43-49 if the given "compound" state- ment is true or false. 43. p and q 47. q and r 44. p or q 48. q or r 45. n and r 49. (p and q) or r 46. por r 50, State the Flat- Plane Postulate in your own words. 51. The following figure is a picture of two distinct "curves" that intersect in two distinct points. State a postulate or theorem that would settle an argument about whether these " curves'" might be lines in our formal geometry. 52. Iff and Q are distinct points and m and n are lines, and if P £ m, Q € ™> P E ?t, and Q £ n, what conclusion can you draw? What postulate or theorem justifies this conclusion? 53. If m and n ai'e distinct lines and P and Q are points, and if m fl n = F and m n n — Q, what can you conclude? What postulate or theorem justifies this conclusion? Review Exercises 39 54. Which postulate assures us that no plane contains all the points of space? 55. Prove, using only the postulates, that every point is contained in at least one plane, 56. Draw a single diagram which illustrates what all eight postulates assert. Label the points in your diagram and then name one pair of skew lines, 57. What dees it mean to say that points A t B, and (7 are collinear? 5S, Does the Three-Point Postulate eliminate the possibility of there being tliree points in a plane that are collinear? 59. Assume that A, B, C, D, and E are five distinct points in the same plane with no three of these points collinear. (a) How many distinct lines are there each containing two of the five points? (b) How many lines determined by A, B, C t D, and E are there on each point? 60. Let A, J3, C, D, and E he five distinct points in the same plane. (a) If A, B, and C are on line 1, and C. D* and F are on line m, and l^m, how many more lines are there each containing two of the five points? How many lines are there altogether? (b) How many distinct lines pass through point A? Point #? Point C? 6L If line I contains three distinct points A, B, and C and if points A and B lie in plane a, what can you conclude? 62. Planes RST and S7T- are distinct. (a) Name a line that lies in both of these planes. (b) Name two more planes determined by points H. S. 7", U. 6,1, tState what intersect means (a) with regard to two lines, (b) with regard to two planes, 64. Which of our Incidence Postulates "pushes" us off a plane into space? Determine in Exercises 65-71 if the given statement is true or false. 65. It is a theorem that every line contains at least two distinct points. 66. The Line- Point Postulate assures us that there is just one line which contains two distinct points. 67* It is a theorem that there arc at least throe distinct lines in every plane, 68. If a line intersects a plane not containing the line, the intersection is a single point 69. It is a theorem that there are at least four distinct planes in space. 70. Our Incidence Postulates assure us of at least twelve distinct lines in space. 71. The Three-Point Postulate assures us that there are at least three dis- tinct points on every line. SydCnrnberg/lXF.t. Separation and Related Concepts 2.1 INTRODUCTION In this chapter we list our second group of postulates. These postu- lates are concerned with points between other points, how a point sep- arates a line, how a line separates a plane, and how a plane separates Space, The ideas these postulates convey arc very simple ones, and the information they give can be easily determined by looking at pictures. We list these postulates so that we can give a meaning to the rela- tionships of bctweenness and separation and to make it quite dear what statements we agree to use in our proof & Later in the chapter we use these relationships to make several definitions. 2.2 THE BETWEENNESS POSTULATES In this section we state the Betwecnness Postulates. They are some- times called order postulates because they tell us how points are ar- ranged in order on a line. For example, if A* B, C are three distinct points on aline, and if Bis between A and C, then we could say that the points are arranged on the line in the Order A* then B, then C We could also say that the order is C, then B t then A, (See Figure 2-1.) a b C -• • 9- Frgrirc 2-1 42 Separation and Related Concepts Chapter 2 In our formal geometry between and order, as they apply to points, are undefined terms. We do not explain by a definition what it means to say that point B is between points A and C Rather we accept the following Betweenncss Postulates as formal statements of betweenness properties. The Betweenness Postulates POSTULATE 9 (The A-B-C Betweenness Postulate) If point B is between points A and C, then point B 1% also between C and A, and all three points are distinct and collinear. Note that the postulate does not allow us to say that "point B is between points A and C" if the points are as shown in Figure 2-2a. Ftpir* 2*1 Why? Similarly, in Figure 2-2b point B is not between points A and C. Why? Figure 2-2c is a correct picture of what we mean when we say "point B is between points A and C." We use the symbol A-B-C or the symbol C-B-A when we wish to say that "point B is between points A and C." POSTULATE 10 (The Three-Point Betweenncss Postulate) U three distinct points are coUincar, then one and only one is between the other two. From this postulate it follows that if A, B t Care three distinct points on a line, then B-A-Q or A-B-C, or B-C-A\ and if A-B-C, for exam- ple, then we cannot have B-A-C or B-C-A, From this postulate it also 2,2 The Betweenness Postulates 43 follows that betweenness for points in our format geometry is different from betweenness on a circle. For example, if three people are seated at a circular table, then most of us would agree that each of them is between the other two. But we should also now agree, after accepting Postulate 10, that betweenness around a table is different from be- tweenness in our formal geometry. POSTULATE 11 {The Line-Building Postulate) If A and B are any two distinct points, then there is | point Xj such that X a is between points A and B, a point Yg Sttdi that B is between A and Y lt and a point Zi such that A is between Zj and B. Figure 2*3 illustrates the meaning of Postulate 1 L We call tliis pos- tulate the Line-Building Postulate because it enables us to prove that there are infinitely many points on a line. Zi a X x B Yi 4 * * • • * ^ Pigut* £-3 From this postulate it follows that there is a point Yi on line AB that is beyond B from A, a point Y 2 that is beyond Y x from B, a point Y 3 that is beyond Yt from Y\, and so on (Figure 2-4). Z\ A Xj B y, F 2 Yi Y± 4 • • • • • * — # — • » » « — ► He..«v v. i Tlius we are able to find on the line as many points as we choose, 3 or 30 or 3,000,000 or any natural numl>er you wish, that arc beyond B from A. Similarly, we can find as many points as we choose that are beyond A from B (Figure 2-5). Z4 ?a Zi Z\ A Xi B Yt Y 2 Y 3 Y. t 4 — • • • • — • — • • • • • • • — • — • — • — •— . ■ ^ Figure 2-5 From this postulate it follows also that no matter how close together two distinct points may be there is always a point between them (Figure 2-6). Z% Z-i Zj A Jfj. X 2 Xs X* Xn B Yi Yi Y S i •• • • • • — * — • • • • » * • • — • m » # | Figure £-6 44 Separation and Related Concepts Chapter 2 Does your intuition tell you that Postulate 11 builds every line com- pletely, that no "holes" could possibly exist in a line? Actually, Postu- late 11 assures 08 that there arc infinitely many points on every line and furthermore, infinitely in any points between every two points of a line. But it does not assure us that there arc no holes in a line. The postula- tional basis for lines is completed in Chapter 3, where we adopt a Ruler Postulate. This postulate provides for the ideas of "continuity' 1 and "infinite extent" EXERCISES 2,2 In Exercises 1-22, indicate by 1" or F whether or not the statement is true or false. In Exercises 1-4, R, S, Tare pointi such that ft-S-T. 1. /i„ $ ? T are collinear points. 2. R. 5, Tare distinct points. 3. T-S-R 4 S-R-T In Exercises 5-KJ, points A and B are on line m-, point C is between A and H and on line n. and m and n arc distinct Hues, 5. m and n intersect at A. G. m and n intersect at B> 7. m and n intersect at C 8. A, B, C are collinear points. 9. A-B-C 10. C-B-A hi Exercises 11—18, F, Q, R are distinct points and no one of them is between the other two. 11. F, Q, R are collinear points. 12. P and Q are collinear points. 13. P and R are collinear points. 14 Q and R are collinear points. 15. There is one and only one plane that contains P, Q, and R, 2 J> The Betweenness Postulates 45 16. There is a point S such that Q-R-S. 17. There Is a point 7 such that Q-T-H, IS. There is a point U such that U-Q-R, In Exercises 10-223 R, S, Tare three distinct eollinear points, R-S-T is false, and 5-R-7" is false, 19. R-T-S 20. S-T-R 2L T-R-S 22. T-S-R Exercises 23-32 refer to Figure 2-7 which shows three noncolhnear points A, B, C and the three lines and the one plane that they determine. Figure 2>7 23. Which Incidence Postulate asserts the existence of three noncoUinear points? 24. Which Incidence Postulate assures us, once wo have points A, B t C, dial lines AB t BC, CA exist? 25. Which Incidence Postulate assures us, once we have points A, B, G\ that there is a plane or containing them? 26. Which postulate assures us. once we have the points and the lines of the figure, that there are points D t Z, F such that B-U-C % C-E-A, A-F-H? 27. Explain why D and E are distinct points. 28. Why is D-E-A a false statement? g& Why is E-A-F a false statement? 30, Which postulate assures us that diere are points G and H such that E-P-C and D-F-IT? 31, Explain why // and C are distinct points. 32, Explain why AB and HG are distinct lines. 46 Separation and Related Concept* Chapter 2 ■ Exercises 33-35. The Incidence Postulates assert the existence of four non- coplanar points. Figure 2-6 suggests four noncoplanar points A, B, C„ D and the four planes and sis lines determined by them. Ffgun»4 33. Deduce from the Incidence and Betweenness Postulates that there is at least one point in space that is not contained in any of the four planes of the figure* 34. Deduce from tl»e postulates that there is at least one line in space that is not contained in any of the four planes of the figure. 35. Deduce from the postulates that there is at least one plane in space that is different from any of the four planes of the figure- Exercises 36-42 refer to Figure 2-9 which shows a line and a portion of it (the heavier part including the points C and D) called a segment (Segment is defined in the next section,} Points X, Y, 2 as well as C and D are points of this line as indicated in the figure. < I £ 1 — I i ► Figure 2-9 36. Are X and V points of the segment? 37. Is Z a point of the segmont? 38. Is C-X-D? 40. Is C-Z-D? 39. Is C-Y-D? 41. Is C C m 42. Try to write a definition of segment. 43. challenge pboblem. Let a line I be given. Deduce from the postu- lates that there are at least three different planes containing /. 44. challenge problem. Let P be u point in a plane a. Deduce from the postulates that there are at least three dLUereuL lines in a passing through P. 2.3 Using Batweenne&s to Make Definitions 47 2.3 USING BETWEENNESS TO MAKE DEFINITIONS We now list several definitions that make use of the between ness relation. Most of these are definitions of terms you have met in your earlier work in geometry. However, it is the betweenness relation that enables us now to make these definitions precise. Figure 2-10 is a pic Lure of a set of points called a segment A B »■ ■ • Figure 2-10 Definition 2,1 If A and B are any two distinct points, seg- ment AS is the set consisting of points A, B, and all points between A and B. The points A and B are called endpoints of AS Sometimes a segment is called a line segment to avoid possible con- fusion with a "segment" of a circle. Be careful not to confuse the sym- bolAi? for a segment with the symbol AB for a line. The horizontal bar in AB reminds us that a segment has endpoints, whereas the bar with * — * arrows in AB reminds us that a line has no endpoints. We sometimes say that two distinct points determine a segment, T hus, if A and B are two distinct points, then the segment determined is AB. Similarly, if A,_B, C are three distinct points, then the segments determined arc AB, BC, and CA. 1 -#- Figure 2- II Figure 2-1 1 is a picture of a set of points called a ray. As the figure indicates, a ray has just the one endpoint A. Speaking informally we might say that the ray continues without end from A through B in a straight line. We express the idea of a ray formally in our next definition. Definition 2.2 If A and B are any two distinct points, ray AM is the union of segment AB and all points X such that A-B-X. The point A is called the endpoint of AB. 48 Separation and Related Concepts Chapter 2 In the symbol AB t the arrow reminds us that it is a ray. The arrow in this symbol is always drawn from left to right regardless of the di- rection in which the ray points. However, do not confuse the symbol AB with the symbol BA. Both represent rays, but they do not represent the same ray, as seen in Figure 2-12, Observe that ray AB has endpoint A, whereas ray BA has endpoint B, A B • •- VtayAB R*y A tytiX Figure 2-12 It is often convenient to speak about opposite rays. Figure 2-13 is a picture of what we mean by opposite rays. The figure suggests the following; definition. AC AB B Figure £-13 Definition 2.3 If A is between B and C, then rays AB and AC are called opposite ruys. The betweenness relation C-A-B in this definition requires oppo- site rays to be collincar. Why? They mast also have a common end- point. Thus the two rays shown in Figure 2-14 are not opposite rays even though they "point in opposite directions." Q — Figure 2-14 2.3 Using Betweenness to Make Definitions 49 Note that the definition of opposite rays is equivalent to saying that two distinct rays are opposite rays if they are colfinear and have the same endpoint. We often use the symbol "opp.AB" for the ray opposite AB. Thus, in Figure 2-13, opp AB = AC since both symbols name the same set of points, namely , the ray that is opposite AB. The following summary should be helpful: If A and B are any two distinct points on a line, they determine the six subsets of the line shown in Figure 2-15, with subsets indicated by heavier marking. Be sure to notice the differences among the symbols AB, AB, AB, and BA. line AB = BA 4 £ fc segment a_ B ray AB 4 4 2- opp AB myBA A oppBA 4— ♦ Figure i-15 A segment has two endpoints and a ray has one endpoint. Some- times we wish to speak about those points of a segment or ray that are not endpoints. Accordingly, we state the following definition . Definition 2.4 The interior of a segment is the set of all points of the segment except its endpoints. The interior of a ray, also called a halftone, is the set of all points of the ray except its endpoint. 50 Separation and Related Concepti Chapter 2 EXERCISES 2.3 L Use the A-B-C Betwecnticss Postulate to prove that every point of AB is in AH. 2. Consider the following definition of a segment AB: AB = (X : ,V £ AH. and X = A, or X = B, or A-X-B). Is this definition equivalent to Definition 2.1? 3- Consider the following definition of a ray AB: If A and H are distinct points, then ray AB consists of point A, point B> all points X such that A-X-B, and all points V such that A-B- X Is this definition equivalent to (Definition 2,2? 4. If A and B are distinct points, what is the intersection of rays AB and BA? 5. If A and B arc distinct points, what is the union of AB and BA? 6. Write an alternate definition of segment in terms of intersecting rays. 7. Can two distinct rays have no print in common? Illustrate with a figure. 8. Can two distinct rays have exactly one point in common? Illustrate. 9. Can two distinct rays have infinitely many points in common? Illustrate. 10. Can the intersection of two distinct rays be a set containing two and only two distinct points? 11. Can the intersection of two distinct rays be a ray? 12. Can two collinear rays have no point in common? 13. Is there a segment with no endpoint? 14. Is there a ray with no endpoint? 15. Is there a segment with no interior point? 16. Is there a ray with no interior point? 17> Is there a segment with exactly one interior point? 18. Is there a ray with exactly one interior point? 19. Is there a segment with exactly two interior points? 20. Is there a ray with exactly two interior points? 2L IF A-B-C, what is the intersection of AH and AC? 22. If A-B-C, what is the union of AB and AC? 2a (f A-B-C, what is the intersection of AB and BC? 24. If A-B-C, what is the union of "KB and B~C? 25. If A-B-C, what is the intersection of AB* and ^B? 26. If A-B-C, what is the union of AB and ~C$? 27. If A-B-C. if S is the Interior of BA t and if T is the interior of BC. what is the intersection of S and 7"? 2.3 Using Betweenness to Mak# Definitions 51 28. If A-B-C, if S is the interior of BA, and if T is the interior of BC, de- scribe the union of S and T. 29. If 5 is the interior of a segment AB, what is the union of S and ABP 30. If S is the interior of a segment AB, what is the intersection of S and AS? 31. If A and B are distinct points of the segment HS, what is the union of A3 and 55? In Exercises 32-41 , a subset of a Hne / is named. Points on line I are indicated in Figure 2-16. Using tills figure, write a simpler name for the set. Exercise 32 has been worked as a sample. t 2 5 C fl £ J? Figure 2-1 32, CE H CE = CE 37. tf? n *S .13, Ci U CE 38. ED U CF 34, Dp fl DA 39. imncF 35. o/^j DF H CF 36, ~CF U 5? 40. offi £? n 41. Mnopp BE CA 42. Prove that there is no segment on a line / which contains every point of line /. (Hint: Let A~S be any segment on line I Show that there is a point on line I that is not in AB,} 43. Iff and Q ore two distinct points on a line, does QP = opp PQ? Explain. 44. How many segments do three distinct points on a line determine? Four distinct points? Five distinct points? (A given set of points determines a segment if the endpoints of the segment are in the given set.) 45. Use your answers to Exercise 44 to predict how many segments six dis- tinct points on a lino determine. Test your prediction by counting. 46. (a} A, B f C t D are four distinct points on a line /. The conjunction of the two sentences (1) and (2) is true if the appropriate symbol ( — , -^, «-*>) is supplied over each letter pair (BA). Copy these statements and supply these symbols. (1) BA contains points C and D, but BA contains neither of these points. (2) D belongs to BA, but C does not. (b) Draw a sketch which shows the order of the four points A t B^QD on I. 47. A % B, C. D are four distinct points on a line /. If C g DB, A £ DB, C £ Bu. A £ BA draw a sketch which shows the order of the four points on /.. 52 Separation and Related Concepts 0-,m:.:-2 48, If FQ is opposite to PR* which one of the three points P, Q t R is between the other two? 40. If opp RQ = RF, which one of the points P, Q, R is between the other two? 2.4 THE CONCEPT OF AN ANGLE The concept of an angle is a fundamental one in mathematics as well as in the practical world of the house builder and the engineer. There are several ways to think of an angle. One way is to think of an angle as two noncollinear rays that have the same endpoint. A second way is to think of an angle as the points on two noncollinear rays that have the same endpoint. In one, an angle is a set of rays. In the second, an angle is a set of points* In the following formal definition we agree to think of an angle as a set of points. Definition 2.5 An angle is the union of two noncollinear rays with the same endpoint. Each of the two rays is called a side of the angle. The common endpoint of the two rays is called the vertex of the angle. Notation. The angle formed by the rays AB and AC is denoted by Z BAG. When using a symbol such as Z BAG it is important that the middle letter denote the vertex of the angle. Since the union of A R and AC is the same as the union of AC and AB, it should be clear that Z BAC = Z CAB, (See Figure 2-17.) It should also be clear that many different points may be used in identifying the same angle. Thus, if AB = aS=AD and AE = A? = aS, as indicated in Figure 2-18, then ZBAE = Z CAE = Z DAE = Z RAF = Z BAG, and so on. Figure 2-1" Figure £-18 If a figure or other information makes clear which rays are the sides of the angle, we can denote Z BAE by simply writing Z A, However, 2.4 The Concept of in Angle 53 if there is more than one angle with vertex A as in Figure 2-19, we would not know which angle is referred to by L A, Figure M It is often convenient to label an angle by indicating a numeral or a lower case letter in its interior. Thus in Figure 2-19 we can write L 1 for Z DAC and Z r for Z BAD. When used, the arcs in the figure indi- cate the sides of the angle and the interior of the angle. Sometimes we speak of the angle determined by two noncollinear segments which have a common endpoint as indicated in Figure 2*20. If A B and BC are the segments, then the unique angle determined by them is ZABC Figure %m Observe in the definition of an angle that the sides of an angle are noncollinear rays and therefore distinct rays. In some books this re- striction is not made, A special case of an angle results when iLs two sides are "coincident." It is called the /em angle. Another special case results if the two sides are distinct but coUinear. In this case the sides of the angle are opposite rays and the angle is called a straight angle. In Figure 2-21, Z ABC is a zero angle and ZDEF is a straight angle. A Figure Ml 54 Separation and Related Concept! Chapter 2 However, because zero angles and straight angles are not needed and it is simpler not to consider them, they have been excluded from the formal definitions in this book. EXERCISES 2,4 1. Copy and complete the following definition: An angle is the [T] of two [7] which have a common ondpoinl but do not lie on the same \t]. In Exercises 2-9, draw a picture to illustrate a set satisfying each of the given descriptions. 2. Two distinct coplanar rays whose union is not an angle. 3. Two angles with the same vertex whose intersection is a ray. 4. Two angles with different vortices whose intersection is a ray. 5. Two angles whose intersection is a segment 6. Two angles whose intersection is a set consisting of cxacdy one point. 7. Two angles whose intersection is a set consisting of exactly two points. 8. Two angles whose intersection is a set consisting of exactly three points. 9. Two angles whose intersection is a set consisting of exactly four points. 10. If A, B t C are three distinct points on AB t if A, D„ E are three distinct points on AD, and if A, B, D are noncollinear points, then the union ol 18 and -A D is an angle which we may indicate as L BAD. Using A, B, C, D, E, write the other possible symbols for this angle. 11. In the following figure , the capital letters denote points and the lower case letters demote angles. Using three capital letters, write another name for each angle denoted by a lower case letter. (a)*=ZQ (d)«=Z[?3 (b) y= im (e) r,= Z[?] (c) %= ZQ (f) w= L\T\ 2.4 Th* Concept of an Angia 55 12. How many angles are determined by three distinct coplanar rays having a common endpoint if no two of the rays are opposite rays? By four dis- tinct rays? By five distinct rays? (A given set of rays determines an angle if the sides of the angle are elements of the given set.) 13. Would your answers to Exercise 12 be different if the rays were not all in the same plane? 14. In the figure below, AB and AC are opposite rays. How many angles do the four rays determine? Name them. 15. If two distinct lines intersect, how many angles are determined? (A given set of lines determines an angle if each side of the angle is con- tained in one of the lines of the given set.) 16. If three distinct coplanar lines; intersect in a common point, how many angles do they determine? 17. Name all the angles determined by the segments shown in the figure. How many are there? Which angles can Ijc named by using only the vertex letter? 18. Using letters in the figure in Exercise 17, name the (a) angle with vertex P in four different ways. (b) ray that is the intersection of APRS and I QRS. (c) ray that is the intersection of APSR and / QSR. (d) segment that is contained in the intersection of Z PQR and Z SRQ, (e) point that is not in FR but is in the intersection of I.RPQ and APRS. (f) three distinct rays contained in the union of Z PRS and Z QRS. 56 Separation and Related Concept* Chapter 2 Exercises 19-30 are informal geometry exercises. The formal development of angle measure appears in later sections. Using Figure 2-22, which shows degrees on a protractor, compute the number named in the exercise. Exer- cise 19 has been worked as a sample. figure 2*2a 19. mlDAF = 55 20. ml BAD 21 mlDAH 22. mlHAE 23. mil AH 24. mllAE 25. mlEAD 2«. ml IAD 27. mllAE + tnlEAD 28. mlDAH - mlllAE 29. ml DAE 30. ml DAE + mlDAF 31. Eveiy triangle is the union of three segments, but not every union of throe segments is a triangle. Write a definition of a triangle that you think will "hold water." (The formal definition of triangle appeals in a later section.) 2.5 THE SEPARATION POSTULATES Tn this section we state three more postulates involving die idea of betweenness, which we call the Separation Postulates. These postulates tell us how a point separates a line, how r a line separates a plane, and how a plane separates space. Although the ideas are very simple, we cannot prove them from the postulates thus far agreed on. In order to facilitate their phrasing, wc introduce the idea of a convex set. Definition 2.6 A set of points is called convex, if for every two points F and Q in the set, the entire segment F^Jis in the set. The null set and every set that contains only one point are also called convex sets. 2.5 The Separation Postulates 57 The definition implies that if P and Q are any two points of a con- vex set T, then PQ C T. A line, a ray* and a segment are examples of convex sets of points as Figure 2-23 suggests. Is a plane a convex set of points? = / + +1 _ — *A FQt=ABm- PQ^Alti- Q Figure £-23 The interior of a triangle is a convex set Figure 2-24 shows two choices. Pi, P* and Q r , Q 2y for points P and 9, respectively, in the in- terior of the triangle and in each case the entire segment PQ is con- tained in the set Is the interior of the circle shown in Figure 2-24 a convex set? (We define interior of triangle, circle, and interior of circle formally in later sections,) Fisure 2-24 However, none of the sets shown in Figure 2-25 is a convex set In each instance it is possible to find points Pand Q such that not all of the segment PQ is contained in the set. J 1 p / Figure 2-25 58 Separation and Related Concepts Chapter 2 Let S be a convex set, T a convex set, and R the intersection of the two sets. If R is the null set or if ft consists of a single point, why is R a convex set? Suppose, then, that R consists of more than one point. Let P and Q be any two distinct points in H. Draw an appropriate figure to suggest the sets S T % R and the points P and Q, Then answer the fol- lowing questions. 1. Why is P in the set S? In the set T? 2. Why is pin the set S? In the set T? 3. Why is P£ in the set S? In the set T? 4. Why is PQ in the set R? 5. Does this prove that R is a convex set? 6. Does this prove the following theorem? THEOREM 2.1 The intersection of two convex sets is a convex Figure 2-26 suggests the first of our Separation Postulates. It shows point A dividing or separating line / into two convex sets T S and T, 4 * • • • ► t B A C D Figure 3-36 No point of / is in both S and T, Point A is the only point on line I that is in neither S nor T, If points B and C are in different sets (that la, if B £ § and C £ T, or if C £ S and R £ T) t then EC contains point A* On the other hand, if points C and D are in the same set (that is, if C £S and D £ S, or C £ T and D £ T ), then CD does not contain point A. We state these ideas formally in our next postulate. POSTULATE 12 (The Line Separation Postulate) Each point A on a line separates the line. The points of the line other than the point A form two distinct sets such that 1. each of the two sets is convex; 2. if two points are in the same set. then A is not between them; 3. if two points are in different sets, then A is between them. 2.5 The S*pa ration Postulates 59 Definition 2.7 Let a line I and a point A on I be given, 1. The two convex sets described in Postdate 12 are called hairline* or sides of point A on line I; A is the endpoint of each of them. 2. If C and D arc two points in one of these sets, we say that C and D are on the same side of A, or that C is on the D-side of A, or that D is on the C-side of A. 3. If B is a point in one of these sets and C is a point in the other set, we say that B and C are on opposite sides of A on line / or that B and G are in the opposite halftones of l determined by the point A. Note that a halfline is a ray with the endpoint omitted. Although it is convenient to think of a halfline as Juicing an endpoint, remember that a halfline does not contain its endpoint. (This use of "have" should not bother yon since you have friends, but you do not contain them.) Note also that opposite hairlines are collineur and that they have the same endpoint. Since there are infinitely many lines in space that contain any given point, it follows that although a point has only two sides on any given line, it has infinitely many sides in a plane or in space. Normally, we do not speak of the sides of a point except when a given line contains the point and we are discussing the sides of the point on that line. Postulate 13 describes how a line separates any plane containing die line. In Figure 2-27, we see that the points of plane « that do not lie on line / are separated into two sets: (1) those points that are on the B-side of line h (2) those points that are on the C-sidc of line /. « Figure 1-27 60 Separation and Related Concepts Chapter 2 No point is in both of these sets and no point of line I is in either of the two sets. If points B and Care in different sets, then 2?C intersects I. On the other hand, if points C and D are in one set, then CD does not intersect line L We state these ideas formally in our next postulate. POSTULATE 13 ( The Plane Separation Postulate) Each line I in a plane separates the plane. The points of the plane other than the points online / form two distinct sets such that 1. each of the two sets is convex; 2. if two points are in the same set, then no point of line I is between them; 3. if two points are in different sets, then there is a point of line I between diem. Definition 2.8 Let a plane a and a line / in a be given. 1. The two convex sets described in Postulate 13 are called halfplanes or sides of / in plane a; l is the edge of each of them. 2. If C and D are two points in one of these sets, then we say that C and D arc on the *ame side of / in plane a t or that C is on the D side of t t or that D is on the C-stde of I or that C and D are in the same h airplane. 3. If B is a point in one of these sets and C is a point in the other set, wc say that B and C are on opposite sides of / in plane a or that B and C are in the opposite halfplanes of a determined by the line L Note that a halfplane does not contain its edge and that opposite halfplanes are coplanar and have the same edge. Furthermore, since there are infinitely many planes in space that contain any given line, it follows thai although a line has only two sides in any given plane, it has infinitely many sides in space. Normally, we do not speak of the sides of a line except when some given plane contains the line and wc arc discussing the sides of the line in that plane. Figure 2-28 suggests the last of the Separation Postulates. The figure shows how a plane a separates the points of space that do not lie in plane a into two sets: (1) those points that arc on the B-side of plane a; (2) those points that are on the C-side of plane a. No point is in both of these sets and no point of plane a is in either of the two sets. If points B and C are in different sets, then BC inter- 2,5 Th* Separation Postulates 61 Figure £28 sects plane a. If points C and D are in the same set, then CD does not intersect plane a. Although these space separation ideas can be proved using the postulates introduced up to this point we shall summarize and state these ideas formally as a postulate in order to simplify and shorten our formal development of geometry. Note the similarity of the Space Separation Postulate to the Plane Separation Postulate. POSTUIATE 14 (Tke Space Separation Postulate) Each plane tit in space separates space, The points in space other than the points in plane a form two distinct sets such that 1. each of the two sets is convex; 2. if two points arc in the same set, then no point of plane a is between them; 3. if two points are in different sets, then there is a point of plane a between them. Definition 2.9 Let a plane a tie given. 1. The two convex sets described in Postulate 14 are called half spaces or aides of plane a and plane a is called the face of each of them. 2. If C and D are any two points in one of these sets, then we say that C and D arc on the same side of a., or that C is on the D-side of a, or that D is on the C-side of a, or that C and jD are in the same halfspaec, 3. If B is a point in one of these sets and C is a point in the other set, then we say that B and C are on opposite sides of a or that B and C are in opposite halfspaces. 62 Separation and Refated Concepts Chapter 2 Note that a halfspace does not contain its face and that opposite halfspaces have the same face. Furthermore, although a point or a line has infinitely many sides in space, a given plane has only two sides in space. The statements of the Separation Postulates and the definitions accompanying them are lengthy. However, the ideas they convey arc simple and can be easily described by means of figures. Briefly, the Separation Postulates tell us that a point separates a line into two half- lines; that a line separates a plane into two halfplanes; that a plane separates space into two halfspaces; and that these sets are convex. Notation. Figure 2-29 shows point A on line I and two halftones de- termined by point A, Wc may denote the halHine on the C-side of A in c ► line / by the symbol U or by the symbol AC. Similarly, the halfline on the B-side of A may be denoted by I2 or by AB. Be careful to note the difference between the symbol AB for ray AB and the symbol AB for hairline AB. Does AB = a3? Explain. p C ■*/ Figur«S- 28 Figure 2-30 shows line I in plane a and the two halfplanes which line I determines. We may denote the halfplane on the /J-side of line 1 in plane a by ffi, and the halfplane on the C-side of line I in plane a by a 2 . Similarly, if S represents the set of all points in space, we may de- note the two halfspaces into which space is separated by a plane with the symbols Si and S2. D« Figure 2-30 2.5 The Separation Postulate* 63 We make use of the Separation Postulates to prove the next two theorems, THEOREM 2,2 If m segment has only one endpoint on a given line, then the entire segment, except for that endpoint, lies in one half plane whose edge is the given line, RESTATEMENT: Given: Line I and segment AB in plane a such that / and AB have only the point A in common, To Prove: If X is any point of AB such that A-X-B, then X is on the B-sade of Mn plane a. Figure 2-31 Proof. 1, X is a point such that A-X-B. 2, A, X, B are distinct points, 3, I does not intersect XK 4, X and B are not on opposite sides of I 5, X and B are on the same side of /, 1. Given 2. Why? 3. Why? 4. Why? 5. Plane Separation Postulate We have shown that X, which is any point of -AB except point A or point B, is on the B-side of t in plane a; hence every point of AB, except point A, is on the B-side of / in plane a. THEORFM 2.3 If the intersection of a line and a ray is the end- point of the ray. then the interior of the ray is contained in one half- plane whose edge is the given line, RESTATEMENT: Given: AB and line / i ntersect in j ust the point A in plane a. To Prove: All points of AB are on the B-side of / in plane a. 64 Separation and Related Concepts Chapter 2 Figure 2-38 Proof: Let X be any point of AB such that A-B-X. 1. All the points of All, except A, lie on the 2Mde of / in plane cl Why? 2, X £ I Why? & I does not intersect BX. Why? 4, X is not on the opposite side of I from B, Why? 5, X is on the same side of t as B, Why? Since X is any point such that A-B-X, it follows that crop BA Ues entirely on the B-side of I Since AB is the union of opp BA and ?0f with A deleted, it follows that A ft is on the B-side of I EXERCISES 2.5 1. Prove that the intersection of any three convex sets is a convex set, 2. There is a theorem which "extends" Theorem 2.1 to any numl>er of sets. Stale this theorem. 3. Is a ray a convex set? 4. Is the interior of a ray ft convex set? 5. Consider the following definition, of a ray: Ray AS consists of point A and all the points on the B-side of A on line AB. Define opp ~AB in a similar manner. 6* Complete the following: Let SI be a segment. Then AB — AB f\ BA. Since AB and BA are convex sets it follows from [7] that [T] is a convex set. 7. Let I be any line and let P and Q be any two distinct points on /. If X is any point between P and (>, is X on line I? Why? Does this prove that PQ is contained in I? Does this prove that every line Is a convex set? 8* Let a be any plane and let Pund Q be any two distinct points in plane «. (a) Is PQ in plane a? Why? (b) Is PQ* subset of ?$* Why? 2.5 The Sepa ration Postulates 65 (c) Is PQ a subset of plane «? Why? (d) Does this prove that every plane is a convex set? 9. Hie interior of each circle shown in the figure below is a convex set (a) Is the intersection of these interiors a convex set? Why? (h) Is l he union of these interiors a convex set? 10, Draw two circles in such a way that the union of their interiors is a convex set, *— • * — » Exercises 1 1-15. In Figure 2-33, lines AB and CD intersect at E so that A-E-B and C-E-D. The D-side of a2 has been shaded. Figure £33 11. Copy the figure and shade the B-side of 12. Why are the two shaded half planes coplnnar? 13. Why is the intersection of these two half planes a convex set? 14. Describe in your own words the intersection of the two shaded half- planes. 15. Does your description in Exercise 14 suggest a definition of "interior of an angle?" 16. Let P, Q, and R be distinct points on a line L with ft and Q in the same hatfline with endpotnl P, On the basis of the given information is it pos- sible that P is between Q and R? That Q is between H and P? That R is between P and Q? 17. Let F, Q, R be distinct points on a line /, with R and Q in opposite half- lines with endpoiut P. On the basis of the given information, is it pos- siWe that JP is between Q and R? That Q is between R and F? That R is between P and OP 66 Separation and Related Concept! Chapter 2 18. Is the union of two opposite halflmes a line? Explain. 19. Is the union of two opposite halflines a convex set? 20. Is the union of two opposite rays a line? Is their union a convex set? 21. In what respect does the set of points in ray AB differ from the set of o— ► points in halflme AB? 22. Is the interior of ray AB the same set of points as halfline a2? 23. Is the u niou of two opposite h airplanes a plane? Ls their union a convex set? 24. Line / lies in plane a, Point A is on one side of / in plane n. Point B is in plane a and is not on the A -side of I and is not on the opposite side of / from A. Make a deduction, 25. A, B, C, X arc lour distinct points on line m, B and A are on opposite sides of A* and C is on the A -side of X. Draw a conclusion about points B and C. 26. E, F y C are three distinct points in plane a. E and F are on opposite sides of line n in plane a. If E and C are on opposite sides of line ft, what con- clusion can you draw with regard to points G and F? 27. From which postulate may we infer that a given line in a given plane Ixas only two sides? 28. Explain why the following statement is true: If J* and Q are any two dis- tinct points in halfplane on, then FQ is in a*. 29. A halfplane is an example of a "connected region," A line in a plane separates the points of the plane not on this line into two connected regions. Into how many distinct connected regions do two distinct inter* seeting lines separate the remaining points of the plane that contains them? 30. Into bow many distinct connected regions do three distinct coplanar lines separate the remaining points of the plane that contains them if no point lies on all three lines and if each two of the lines intersect? Zh Into how many distinct connected regions do four distinct coplanar lines separate the remaining points of the plane that contains them if no three of the lines contain the same points and if each two of Lhe lines intersect? 32. Use your answers to Exercises 29-31 to predict the number of distinct connected regions into which five distinct coplanar lines separate the remaining points of the piano that contains them if no three of the lines contain the same point and if each two of the lines intersect 33. Can three distinct coplanar lines be situated so as to separate the re- maining points of the plane that contains them into three distinct con- nected regions? Four distinct connected regions? Five distinct con- nected regions? Six distinct connected regions? Seven distinct connected regions? More than seven distinct connected regions? 2,6 Interiors and Exteriors of Angles €7 Figure 2-.M 34. Draw a figu re for each part of Exercise 33 to which yon answered "Yes. 35. Figure 2-34 shows two distinct planes ft and /? intersecting in a line t The line Z is the edge of taw many different halfplanes represented in the figure? 36. Name two distinct halfplanes represented in Figure 2-34 that are coplanar. 37. Name two distinct halfplanes represented in Figure 2-34 that are not coplanar. 38. How many different pairs of halfplanes in Figure 2-34 are not coplanar? 39. Chic plane separates the rest of space into two connected regions. Into bow many distinct connected rejpons do two distinct intersecting planes separate the rest of space? 40. Into how many distinct connected regions do three distinct intersecting planes separate the rest of space if no line lies in all three of the planes, if every two of the planes intersect, and if each plane intersects the line of intersection of the other two pla 41. L'se your answers to Exercises 39 and 40 to predict the number of dis- tinct connected regions formed by four distinct intersecting planes if no three of these planes contain the same line, if each two of these planes intersect, and if each plane intersects each line of intersection formed by two of the other planes. 42. challenge problem. Construct a model to represent the situation of Exercise 41. Count the number of distinct connected regions formed. How does this number compare with your prediction? 43. challenge problem. Extend the result of Exercise 41 to five planes. 2.6 INTERIORS AND EXTERIORS OF ANGLES In Section 2.5 we introduced the concept of separation. A line in a plane separates the points of the plane not on the line into two half- planes. A pie rure of an angle suggests that an angle separates its plane. Indeed, if plane a contains tLABC, then all the points of at that are not points of L A BC make up two sets, one called the interior and the other the exterior of Z ABC. We shall state carefully what we mean by these terms. Separation and Related Concepts Chapter 2 One of the simplest ways ly think of the interior of an angle is as the intersection of two halfplanes associated with the angle, For /.ABC %ur*2-35 these haifplanes are the C-sitle of AS and the Aside of BC, as indicated in Figure 2-35. Our formal definition is as follows: Definition 2.10 The interior of an angle, sa y £ABC t is the intersection of two haifplanes, the C-side of AB and the A-side of BC. Figure £38 Definition 2.11 The exterior of an angle is the set of all points in the plane of the angle except those points on the sides of the angle and in its interior. Figure 2-36 illustrates both Definitions 2. 10 and 2.1 1. Tile following theorem is easy to prove using Definition 2.10 and Theorem 2.3, 2.6 Interiors and Exteriors of Angles 69 THEOREM 2.4 If P is any point in the interior of Z ABC, then the interior points of ray BP are points of the interior of Z ABC RESTATEMENT: Given: Z ABC with P a point in the interior of Z ABC. To Prove; B?is in the interior of /.ABC Proof: By definition of the interior of L ABC, P is on the C-side of AJB and on the A-side of BC. By Theorem 2.3, BP is on the C-side of AB * — » o • and on the A -side of BC. Therefore BP is in the intersection of these two halfplanes which, by definition, is the interior of /ABC. c Figure 2-37 EXKRCISE5 2.6 L Copy and complete the following definition. The interior of /PQR is the [7j of the haffplane lhatis the P-side of [f] and the halfplane lhat is the \?j oi PQ. Exercises 2-7 refer to Figure 2-38. 2* Which of the labeled points are in the interior of Z ABC? & Which of the labeled points arc in the exterior of Z ABC? 4. Which of the labeled points are not in the interior of L ABC? 5. Which of the labeled points are in the interior of Z GBR? 6. Which of the labeled points are in the interior of Z G7J/JP 7. Which of the labeled points are in the interior of Z CBG? 70 Separation and Related Concepts Chapttr 2 §, Is the vertex of an angle u point of the interior of the angle? Explain. 9. Is the vertex of an angle a point of the exterior of the angle? Explain, 10. Is B a point of the interior of Z ABC? Explain. 11. Is B a point of the exterior of LABC? Explain. 12. Suppose that A, B, C, D are four noncoplanar points. Is it possible that D k a point of the interior of <LABC? Explain. 13. Suppose that A, B t C, D are four noncoplanar points. Is it possible that D is a point of the exterior of £ABC? Espial ■ Exercises 14-20 refer to Figure 2*39. Kipire 2-3S 14. Make a copy of Figure 2-39 and shade the A^side of F.B with vertical raw ff and the £-sfde of AB with horizontal rays ^, 15. Which angle has only vertical shading in its interior? 16. Which angle has only horizontal shading in its interior? 17. Which angle has both kinds of shading in its interior? IS, Which angle has no shading in its interior? 19. Is the intersection of the I wo shaded portions the interior of any of the angles shown in the figure? If so, name the angle(s). 20. Is the union of the two shaded portions the exterior of any of the angles shown in the figure? The interior? 21. Does au angle separate the points of its plane not on the angle into two connected regions that do not intersect? 22. What axe the connected regions of Exercise 21 called? 23. Draw a picture of a n angle. Mark two points Pand Q hi the exterior of the angle. Does PQ intersect the angle? 24. Does your answer in Exercise 23 depend on your choice of P and Q or would the answer he t he same for every choice of P and Q? 25. Is the exterior of an angle a convex set? 26. Draw a picture of an angle. Mark two points P and Q such thai P is in the interior of the angle and Q is in the exterior. Does PQ intersect the angle? 2,6 Interiors met Exteriors of Angles 71 27. Does your answer in Exercise 26 depend on your choice of F and Q t or would the answer be the same for every choice of P and Q if P is chosen in the interior of the angle and Q in the exterior? 28. challenge PimiiLEM. It can he proved that if P is a point in the in- terior of an angle and Q is a point in the exterior,, then PQ intersects the angle. The proof is difficult and there are essentially five cases to con- sider as indicated in Figure 2-40 where £ ABC is the angle, P is a point in the interior of the angle, and Qi, Qs, Q& Q± t Q& are points in the ex- terior of the angle such that ■i — * *— i ■» (1) Qi is any point on the A -side of BC and on the opposite side of AS from C. (2) Q 2 is any point on Vpp B( ^ y (3) Q3 is any point on the opposite side of AB from C and on the oppo- site side of BC from A* {4} Qa is any point on &pp BA, (5) Qs, is any point on the C-side of AB and on the opposite side of BC from A. * / ■% Wjjure 240 Following is an outline of a proof for case 1, a proof that if Qi is any point on the A-xidc of BC and on the opposite side of AB from C, then ~F^fi intersects LABC. Answer the "Whys" in this outline, i. P and C are on the same side of A H. Why? 2. C and Qi are on opposite sides of AB. Why? 3. P and Qi are on opposite sides of AB* Why? i — » 4. There is a point R between P and Qi on line BA, Why? <■ — * This means that PQi intersects AB in an interior point of PQi . We wish now to prove that PQ\ intersects BA rather than Qpp BA. 72 Separation and Ralated Concepts Chapter 2 5. P Is oil the A-side of M Why? 6. Q t is on the A-side of K?. Why? 7. The A-side of BC is a convex set Why? B. AH points of FQx are on the A-side of KJ. Why? Therefore the point fl is on the A-side of BC and on M. Therefore R is on BA, and hence H is a point of I ABC. Therefore PQi intersects labc. 29. challenge pboblkm . Let the situation of Exercise 28 be given. Prove that PQ 2 intersectsTJA. 30. CHALLENGE problem. Let the situation of Exercise 28 be given. Prove that P{) s intersects L ABC. 31. Draw a picture of a n angle. Mark two points F and Q in the interior of the angle. Does PQ intersect the angle? 32. Docs your answer in Exercise 31 depend on your particular choice of P and Q? 33. Is the interior of an angle a convex set? 34. challenge pkoblem. Use Theorem 2.1, the Plane Separation Postu- late, and the definition of interior of an angle to prove that the interior of an angle is a convex set. 35. On the basis of your experiences in informal geometry try to write a definition of a triangle, thinking of it as a set of points. (Remember that a segment is a set of points and that the union of several sets of points is a set of points.) 36. On the basis of your experiences in informal geometry try to write a definition of a quadrilateral. 2.7 TRIANGLES AND QUADRILATERALS Next to segments and angles perhaps the simplest geometrical fig- ures are the polygons, and the simplest polygons are the triangles. You all know what a triangle looks like. It has three sides and three angles, A drawing of a triangle (Figure 2-11) shows its three sides, which are the segments AB, BC, and GA. 2.7 Trianghw and Quadrilaterals 73 On the other hand, the angles of a triangle are not shown completely in a picture of the triangle. In Figure 2-42, however, there are pictures showing the tingles of a triangle. A picture that does show all the angles of a triangle is not a picture of what is usually meant by a triangle. its uigto* Figure 2-42 The following definition is worded carefully so that it says exactly what we want it to sav. Definition 2.12 If A, B f C are three noncollinear points, then the union of the segments A B, EC, CA is a triangle. Notation. The triangle which is the union of the segments AB, BC, CA is denoted by AABC x\n alternate definition which you might prefer is as follows. ► A triangle fe the union of the three segments determined by three noncollinear points. This is an acceptable definition since we have agreed (in Section 2.3) that "if A, B. C are three distinct points, then the segments de- termined are AB, BC, and CA." 74 Separation and Related Concepts Chapter 2 Definition 2J3 Let A AiJC be given. 1. Each of the points A, B. C is a vertex of A ABC 2. Each of the segments AB, W, CA is aside of A ABC. 3. Each of the angles L ABQ Z BCA, L CAB is an an^e of AABC. 4. A side and a vertex not on that side are opposite to each other. 5. A side and an angle arc opposite to each other if that side and the vertex of that angle are opposite to each other. Note that a triangle contains its vertices and its sides but that it does not contain its angles. An angle of a triangle is not a subset of the tri- angle. It is customary to speak of the "angles of a triangle" or "the angles determined by a triangle," but it is incorrect to refer to them as "the angles contained in the triangle." Remember that whereas a tri- angle has angles, it does not contain them. (In this connection it might be helpful to think of a farmer who has farms and barns but does not contain them, or of a man who has a car and a house and lot but does not contain them!) Figure 2-43 illustrates the interior of a triangle and the interiors of the angles of a triangle* AA HC, l~A t interior of L A AABC, jLB, interior of Z_B AABC, LC, interior of £C Figure £-43 AABC, interior AABC 2.7 Triangle* and Quadrilateral* 75 Definition 2.14 The intersection of the interiors of the three angles of a triangle is the interior of the triangle. Definition 2.15 The exterior of a triangle is the set of all points in the plane of the triangle that are neither points of the triangle nor points of the interior of the triangle. In Figure 2-44 the interiors of Z A and L B are shaded. Note that the intersection of these interiors is the interior of the triangle. figure 2-M This Suggests the following theorem, THEOREM 2. 5 The i ntersecrion of I he interiors of t wo angles of u triangle is the interior of the triangle. Proof: Let A ABC be given. I^et X be any point in the intersection of the interiors of any two of its angles, say L A and Z B. X is on the B-side of AC because it is in the interior of LA. X is on the A -side of BC because it is in the interior of Z B. X is in the interior of Z C because it is on die B-side of AC and the A-side of BC. This proves that if a point is in the interiors of two angles of a tri- angle, then it is also in the interior of the third angle. Therefore the in- tersection of the interiors of two angles of a triangle is contained in the 76 Separation and Related Concepts Chapter 2 interior of the triangle, Since the interior of the triangle is the inter- section of the interiors of all three angles, it follows that the interior of the triangle is contained in the intersection of the interiors of any two of its angles. Therefore the interior of the triangle is contained in, and also contains, the intersection of the interiors of any two of its angles. Therefore the interior of the triangle is the intersection of the interiors of any two of its angles. Consider any three n on colli near points A, B, C in a plane a and a line I in a which does not contain A or B or C but which does contain an interior point of AC as shown in Figure 2-4*5. Figure fc-45 L B is either on the C-side of I or on the A-side of / in plane a. Why? 2. If B is on the C-sidc of I, then I intersects AB in a point in the interior of AB, Why? Does I intersect BC in this case? Why? 3. If B is on the A-side of /, then I intersects BT in a point in the interior of BC. Why? Does / intersect AB in this case? Why? 4. Does this prove the following theorem? THEOREM 2,6 If a line and a triangle are coplanar, if the hue does not contain a vertex of the triangle, and if the line intersects one side of the triangle, then it also intersects just one of the other two sides. Triangles have three sides. Quadrilaterals have four sides. By quad- rilaterals we mean some., but not all, plane figures that are made up of four segments as suggested by Figure 2-46. Try writing a definition of quadrilateral before reading further. Then compare your definition with the following definition which we adopt for our formal geometry. 2.7 Triangles and Quadrilaterals 77 Quadrilateral* Not quadrilftt*rjiln Figure 2-46 Definition 2. 16 Let A, B T C, D be four coplanar points such that no three of them are collinear and such that none of the segments AB, ffll, CD, DA intersects any other at a point which is not one of its endpoints. Then the union of the four segments AB, B£, UD, DA is a quadrilateral. Each of the four se^nents is a side of the quadrilateral and each of the points A, B, C, D is a vertex of the quadrilateral. Is it possible that four given points are the vertices of more than one quadrilateral? Figure 2-47 shows that this is indeed possible. Think of the figure as four different pictures of the same four points. The sec- ond, third, and fourth pictures show different quadrilaterals with the same vertices. p A E ASDC a urn ADBC Figure 2-47 To name a quadrilateral using the names of its vertices, and to do it so that we know which segments are its sides, the names of the ver- tices are so arranged that { 1) letters adjacent to each other in the name 78 Separation and Related Concepts Chapter 2 of the quadrilateral are names of the endpoints of a side of the quadri- lateral and (2) the first and last letters in the name of the quadrilateral arc names of the endpoints of a side of the quadrilateral. Note that if AB CD is a quadrilateral, then BCDA is the same quad- rilateral. Give several additional names for this quadrilateral. Notice also that ABCD and ACBD are not names for the same quadrilateral. EXERCISES 2.7 1. Copy and complete the definition: If A, B, C are three [?] points, then is the triangle denoted by A ABC. 2. Name the side which is opposite lo /. ABC in A ABC. % Name the angle which is opposite to side AB in A AUG'. In Exercises 4-10, let A ABC be given. State whether or not the given set is a subset of A ABC, 4. AABC 8. AB U BC J CA 5. Interior af / ABC 9. AB f BC ft 25 6. Interior of AABC 10. {A, B s C} 7. AB In Exercises 11-1-5, let AABC be given. Let S denote the union of AABC and its interior. State whether or not the given set is a subset of S. 11. Interior of £ABC 12. (Interior_of_/ ABQ D (Interior of IBCA) 13. [AB.BC, CA) 14. AB HBCnCA 15. AB n (Interior of A ABC) In Exercises 16-19, let AABC be given. Let II U H 2x H 3 denote the follow- ing halfplancs, respectively: the A -side of BC, the B-side of CA, and the C-side of AB. Express the given set in terms of these halfplanes. 16. Interior of £ABC 18. Interior of ZCAB 17. Interior of LBCA 19, Interior of AABC In Exercises 20-24, let the same situation as in Exercises 16-19 be given. State whether or not the given set is a subset of the exterior of AABC. 20. Tlie interior of the ray 22. BC opposite to BC 23, BC 21. The ray opposite to BC 24. The opposite halfplane front //j 2.7 Triangles and Quadrilaterals 79 25. Draw a picture of a triangle and a line so lhat (a) their intersection is one point; (b) their intersection is exactly two points; (e) their intersection consists of more than two points. 26. Gun a line intersect a triangle in exactly one point which is not a vertex? Illustrate. 27. What is your answer to lixercise 26 if the line and triangle arc contained in the same plane? 28. Can a line intersect a triangle in exactly three points? 29. Name all of the triangles shown in the figure. D 30. If A , B t C are the vertices of a triangle, prove that there is a point of L A that is not a point of the triangle. 31. If A, H f C are noncollincar points, is the following statement true? AABC = {I A n IB) U {I A n Zqj(ZBn ZC> 32. Let A ABC with points P and {> such that A-F-B and A -Q-C as in the figure be given. To prove that A ABC is not a convex set of points, it is sufficient to show that there is a point X of fQ which is not contained in A ABC. Let X be a point of TQ such that P-JE-Q, A (a) X £ AC. (b) X € AB. (c) X £ BC. Why? {Hint: Use Tlicorem 2.6.) Does this prove that B triangle is not B convex set? 33. Is the exterior of a triangle a convex set? Illustrate with a figure. 34, Is the interior of a hi angle a convex set? Kxplain why. 80 Separation and Related Concepts Chapter 2 ■ For Exercises 35-38 draw a triangle, A ABC. 35. Mark a point that is in the interior of Z A and in the exterior of the triangle. 30* Mark a point that is in the interior of Z A but is neither In the interior nor the exterior of the triangle. 37. Mark a point in the exterior of the triangle that is not in the interior of any of the angles of the triangle. 38, Can you mark a point that is in the interior of Z A and is in the interior of ZB but is not in the interior of ZC? Why? ■ In Exercises 39-42, let quadrilateral ABCD be given. 39, Is it possible that D is an element of BC? 40, Is it possible that D is an element of the interior of L BAC? 41. Is it possible that D is an element of the exterior of Z BAC? 42. Is it possible that the interior of A ABC is a subset of the interior of A A DC? 43. Prove that every plane contains a quadrilateral, 44, Prove that any three vertices of a quadrilateral are the vertices of a triangle. 2.8 PROPERTIES OF EQUALITY AND NUMBER OPERATIONS In preparation f or Chapter 3, which involves some algebra, we con- chide Chapter 2 with several sections devoted to a review of elemen- tary algebra. Chapter 3 contains many equations, that is, statements of equality. Following arc some basic properties of equality and num- ber operations that are useful in working with equations involving numbers. Substitution Property of Equality. In any statement about some thing (physical object, number, point, line, idea, etc) one of the names for that thing may be replaced by another name for the same thing. If the original statement is true, then the statement with the substitution made is also true. If the original statement is false, then the statement with the substitution made is false. Recall that A = B means that A and B are names for the same thing. The substitution property tells us that we may replace A by B in any statement about A without changing the truth or falsity of the statement. 2,8 Equality and Number Operations 81 Example Consider these two statements: (1) 7 = 2 + 5 (2) 3 + 4 = 7 From (1) and (2) it follows by the substitution property of equality 3+4 = 2 + 5. Reflexive Property of Equality. For any thing a, a = n. Symmetric Properly of Equality. For any thing a and for any thing b t if a = h, then h = a. Transitive Property of Equality. For any things a, b, c, if a = b and b = c, then a = c Addition Property of Equality for Numbers. If a, h> c are real num- bers, and if a = h y then a + c = b + c. More generally, if a, fc T c, d are numbers such that a = b and c = d* then a + c = fr + d and a — c= b — d. (It is appropriate to include subtraction here since each difference can be expressed as an addition. Thus a — c — a + (— c). Also if c = d, then — c = —d.) Multiplication Property of Equality for Numbers. If a, b, c arc real numbers and if a = b t then ac = be. More generally, if a, b> c t d arc numbers, and if a = /; and c = d, then ac = bd t and if i c=£0,d=£Q f then c (It is appropriate to include division here since each quotient can be expressed as a product . Thus — = a* — - Also, if c = d* c =£■ 0, c c <f=^0\ then 1 = 1) c d 82 Separation and Related Concepts Chapter 2 Commutative Property of Addition. If a and b are real numbers, then a -f b m b 4* a. Commutative Property of Multiplication. Tf a and h arc real num- bers, then ab = ha. Associative Property of Addition. If a, h t c are real numbers, then (a + h) + c = a + {h + c). Associative Property of Multiplication, If a z /?, c are real numbers, then [ab)0 = a(bc). Distributive Property of Multiplication over Addition, If a, h, c are real numbers, then a(b + c) = ab + ac and [a + b)c = ac 4- fee, EXERCTSES 2.8 ■ In Exercises l-20 t name the property that justifies the given statement L tf x = 5 and x = y. then, y = 5. 2, If ac = 5 and 5 = y, then x = y, a 7 = 7 4. If AB = CD, then CD = AB. 5. If 3 = x and 4 = y, then 3 + 4 = x 4- y. 6* if 3 = x and 4 = y, then 12 = xy, 7. If § is a coordinate system, then $ = g. 8. If x 4- y = z and if x = a + h, then (a + b) 4- y a ft 9. If lit = tcandif a 4- fr = uandc 4- d 1 = c, then (a 4- fr)(c -f tt) = 10. 10. If -£ = c, tiien a = c/j. o IL Ifa + fc=c, then (o 4- fc) 4- (-&) = c + (-b). 12. If f";? _ 6 lhcn x _ 5 _ ^ _ Q j 13. 3(20 + 5} = 3 • 20 4- 3 ■ 5 14. 3(xy) = (3% 15. (2 4- 3)5 = 2 • 5 + 3 • 5 16. (2 4- 3){4 4- 7) = (2 4- 3)4 4- (2 4- 3)7 17. (3 + 2-3} + 7 = 3 4- (2-3 4- 7) 2.9 Sohrfmg Equation* 83 IS. 5 + i=i + 5 19. 3 + (4 + 7) = 3 + (7 + 4) 20. 3{7 + x) = (7 + x)3 2.9 SOLVING EQUATIONS The properties of equality and number operations are useful in solv- ing elementary equations. Example 1 Solve for xt x + 3 = 52. Solution: x + 3 = 52 Given x = 49 Addition property (Acid -3) Example 2 Solve for *; 3,t - 5 = 7x + 2. So?u£ion: &r - 5 = 7x + 2 Given 3:t = 7* + 7 Addition property (Add 5) -4x= 7 Addition property (Add — 7x) i= -I 4 Multiplication properly (Mult by -i) Example 3 Solve for x - 5 x + 8 10 12 Solutions r — a _ a: + 10 12 Given 12(x - 5) = J0{x + 8) Multiplication properly (Mull. by 120) 12r — 60 = 10* + 80 Distributive property 12* = lOt + 140 Addition property (Add 60) 2* - 140 Addition propei+y (Add — lOx) * = 70 Multiplication property ('Mult. by^> 84 Separation and Related Concepts Chapter 2 Example 4 Solve for x\ X — o x-7 (**T) Solution: X-5 _ 5 *- 7 " 8 8(x - 5) = 5(x - 7) fix - 40 = 5* - 35 3* =5 X = 5 Given Multiplication property (Mult by8(*-7)) Distributive property Addition property (Add ( - 5x + 40)) Multiplication property (Mult, JErompte 5 Solve for « x + ° = * * + 5x. ^ o Solution; Jl±A = JLzi1+5x 2 3 3(* + 5) = 2(x - 1) + 30x + 15 = 2x - 2 + 30x -29*= -17 x = XL 29 Given Multiplication property (Mult, by 6} Distributive property Addition property (Add (-32a: - 15)) Multiplication property (Mult. by -i) Although the list of properties does not explicitly include the divi- sion and subtraction properties of equably for numbers, they are in- cluded implicitly. For example, dividing by 7 is the same as multiplying by \, and subtracting 5 is the same as adding —5, In Example 1 you could think of subtracting 3 and justifying it by the subtraction prop- erty instead of adding — 3 and justifying it by the addition property. In Example 2 you could obtain the last step from the preceding step by dividing by —4 and justifying it by the division property instead of multiplying by — j and justifying it with the multiplication property, Note, however, that the multiplication property does not permit us to divide by 0, since is not the reciprocal of any number. 2.9 Solving Equations Example 6 Solve for x: a(x + x t ) = 5, where a =£ 0. Solution: a(x + «J = 5 * + x x - ^ a x = — _ Xl Given Division property (Divide by a) Subtraction property (Subtract EXERCISES 2.9 In Exercises 1-20, solve for x. Name the properties that justify the steps in your solution. Express your answer in simplest form. 1. 3x + 3 = 2x + 4 2. 3s* - 4 - 4x + 3 3. i* = J*+i 4. 0,75* - 100 5. 4(x - 64) = 28 i±4£«a 5 - 3(*~1) _. 1 ' s s " 8-5 -9 -(-6) ft x ~ l - 'J ~ 3 * 7 - I " 15 - 3 a: - 10 fc - () 11, 4x - 5 = 14c - 75 _1_3 - _]0-(-8j ia x 4- 4 = 40 - * *-7 „ 7 -(-3) 14. 10. -11 - 10 5- 1 1 15. 3{x + 15) = 2{x + 15) 16. 3(.v + 15) + 2{ X + 16) 17 '7^ocH^ 100 > ia ~5— = 7 + '^0^ 20. 0.3x 4- 0.8x a 220 1 - In Exercises 21 -40, solve for x. Express answers in simple form. 21. 3* + 5 BS 22. 3(x- 1) + 2(* - -1) = 4* 23, 5* + 7 = 3{* + 4)- 5 24. x — 2 1 _ x-3 3 25. 2ar- 3 5x- 1 26. *- 1 |-£*** 27. ox -1=4 2S. g- 1 1 x- 1 I* ~ ? 5a:— 1 7-3 x-2 3 86 Separation and Related Concept* 30. 5x ~ 1 = ^nl 6* - 7 6x - 8 31. * + % =3 32. 3x + y = 34. x-3 = Jk + 4 35. 5( j - 3) = 3{* + 4) Chapter 2 36. 5* - 3 = 3* + 4 37,*~ k + 2 2 3 38. 4x - 5 = 3«i + 4xa x — *i 5 39. *2- ari *+! = 7 -5 2.10 EQUIVALENT EQUATIONS In solving an equation, wc find the number (or numbers) that "satisfies" the equation. Such a number is a root of the equation. In Example 1 of Section 2.9 the root of the equation x + 3 = 52 is 49. Note that 49 satisfies the equation since 49 + 3 as 52 is a true sentence and that no other number satisfies the equation. The set of all roots of the equation x 4- 3 = 52 is since 49 is the one and only root of the equation, {49} is the solution set of the equation x 4- 3 = 52. In each example of Section 2.9 we used properties of equality and number operations to obtain other equations that have the same solu- tion set Equations that have the same solution set are called equivalent equations. Note in Example 2, for instance, that anc 3x -5 = 7* + 2 3ac = 7x + 7 -4x - 7, x — 7 4 are four equivalent equations. The solution set of each of them is HI- Sometimes the properties of equality and number operations are used to produce equations not equivalent to the given equation. 2.10 Equivalent Equations 87 Example I x = 5 Given 0=0 Multiplication property (Mult, by 0) It is true that if x = 5, then = 0. Indeed = is a true sentence regardless of what may be true about x. The solution set of x = 5 is obviously {5}. The solution set of = is the set of all real numbers. Even though it is true that if x = 5, then = 0, it is not true that x = 5 and = are equivalent equations. When we went from x = 5 to = 0, we went from an equation with one root to an equation with infinitely many roots. We did not lose any roots, but we certainly gained many of them I Example 2 ■■«=T +1 x(x- 2J= {*- 2) + l(x-2) x* - 2* = x- 2 + x- 2 x 2 - Ix = 2x - 4 x 2 - 4x + 4 = (x - 2)2 = x~2 = x = 2 In this example we gained a root somewhere along the way. Trie solution set of x = 2 is obviously {2}> but 2 is not a root of the original equation. Why? The solution set of the given equation is the null set. If we try to reverse the steps in this "solution/' we can Justify each step except the last one. Given x = 2. we cannot justify going from x{x - 2) = (x - 2) + l(x - 2) to * = 1- -f 1 by dividing by x — 2, The multiplication property permits us to divide by any number except and x — 2 = if x = 2. 88 Separation and Related Concepts Chapter 2 Example 3 x 2 = 3x * = 3 In this example we multiplied both sides of x 2 = 3* by — (or divided both sides by x). In doing this we lost a root. How did this happen? Di- viding by x is legal except if x = 0. Is a root of the given equation? Is 2 = 3 • 0? Yes, it is. The solution set of the given equation Is {0, 3}. The solution set of x = 3 is {3}. The equations x 2 = 3* and x = 3 are not equivalent equations. Tn solving BQ equation through a seqin.-:iu.-i' <j| equations it is advis- able to check each step for a possible loss of roots. If there is a value of x that might be lost as a root (as is lost in going from x 2 = 3x to x = 3 in Example 3), it should be identified and checked by substitution in the original equation. Of course, it is a root if and only if it satisfies the original equation. To make sure that no roots are gained in the solution process you may (I) check each root of the final equation by substitution in the orig- inal equation or (2) check to see if the equations can be obtained in re- verse order without a loss of roots at any step. If there is no loss going backward, there is no gain going forward. If a root is lost going back- ward, it was gained going forward, and hence is not a root of the given equation. In Example 2 we have a sequence of equations and there is no loss of roots in going forward. When we reverse the steps, there is no loss of roots at any step until the last one. 2 is a root of x(x - 2) = x - 2 + l(x - 2) x — 2 but not a root of x = + 1< x — 2 2 is lost as a root going backward; it was gained as a root going forward. Example 4 1 = I Given X — i x — 1 = x - 7 Multiply by x — 7 0=0 Subtract x — 7 In this example we have a sequence of three equations and the so- lution set of the last one is the set of all real numbers. In reversing the 2.10 Equivalent Equations 89 steps we do not lose any roots until the last step. To get x — 7 from x — 7 = x — 7 we divide both sides by x — 7, and this is not permissible if x s= 7. Checking, we see that 7 is not a root of the given equation. The solution set of the given equation is the set of all real numbers except 7. Example 5 n\ y — 2 _ _ x — 2 (2) y - 2 = 7(* - 2) In this example there are two variables, x and y. In Equation (2) we may wish to think of x as the independent variable and y as the de- pendent variable. A solution f Equation (2) is an ordered pair of num- bers {a, b) such that Equation (2) is satisfied when x is replaced by a and y is replaced by b. Thus (3, 9) is a solution of Equation (2) since 9 - 2 = 7(3 - 2) is a true sentence. If we solve Equation (2) for gf, we get y = 2 + 7(* - 2) or y = 7x - 12. Let a be any real number whatsoever. Then x = a and y = la — 12 satisfy Equation (2). For upon substituting a for x and la - 12 for y in (2) we get (la - 12) - 2 = 7(0 - 2), which is a true sentence. Indeed, the set of all ordered pairs (a, la — 12), where a is a real number, is the solution set of Equation (2). Any other letter can be used for the symbol a here. Thus we could say, if we wish, that the solution set of (2) is the set of all ordered pairs (x> 7x — 12), where % is real. Is the set of all ordered pairs fa la - 12) also the solution set of Equation (1)? Let us check Substituting a for x and la - 12 for y in Equation (1), we get (7a - 12) - 2 - a -2 or after simplifying, ' ( fl ~ ) = 1. This is a true statement for every a a — 2 90 Separation and Related Concepts Chapter 2 with one exception. It is not true if a = 2, The solution set of (1) is the set of all ordered pairs (a, la — 12) for a ^ 2, Equations (1) and (2) are not equivalent Some of you have graphed equations like (2). You know that its graph is a line. What is the graph of (1)? The graph of (1) includes all the points of the graph of (2) except the point with abscissa 2, that is, die point (2, 2). The graph of (!) is the union of two opposite halflincs. Or to use a bit of informal language, a line with a hole in it. EXERCISES 2.10 In Exercises 1-20, check by substitution to see if the given value of x is root of the given equation, 1. X 2 + 5x + 6.25 = 0, x = 1 2. a* + 5x + 6.25 = t x= -2 3. x* 4- 5* 4- 6.25 = 0, t - -2.5 4. 3(x - 5) + 7(* - 5) = 10(x - 5), x = 5 5. 3(x - 5) + 7(x - 5) = 10(x- 5), x = 5.1 6. 3(x - 5) + 7(x- 5) = 10(x -5),i= 137.3 7 x - 2 _ 2x + 3 T _ , " 6x~T9* = 2xJ4 2 St- ■ 6 ar — 2 3x- -6 x — 2 3x- -6 x — 2 3x- -6 1 Gx + 9 2*4-3 fix - 9 x= -1.5 10. -*-=-% = -^4- * = 1°°° Gx + 9 ]£ _J_ = _^-,*=2 x 4- 2 a : - 2 x ■+■ 1 14. 1 + x + 1 1 16. x + 1 214-3 x - 3 x — 1 x a - l 1 2x X — 1 x a - l 1 2x X*- X — 1 1 7- - 5 x = - -6 4-— l — = -^— t x= -1 2.10 Equivalent Equation* 91 x — 2 * + 3 i-4 19.^+-*i| + -*^i=3.x=4 x-2 T x + 3 x- 4 2a x^ + JL ^3 x-4 = _ 3 x — 2 x + 3 x— 4 In Exercises 21-30, check to see if the two given equations are equivalent. If they are not equivalent, state whether there is a loss or gain of roots in going from the first equation to the second one. 21. x* = -2x, x = -2 22. 2x=3,x = 1.5 23. x = 5, x* = 25 24. x = 5, x2 = Sx 25, x = 5, x - - 1 =4 26. x = 5,2x ■ I = 11 27, 3x + 2x = : 6x, X = 28. x-2 3 x- 3 4 4(x -2) 29. x-2 3 x - 3 x - x — -2 3 " 3 4 J 1 30 x-2 x-3 X — 3 4 3 4 ' X — 2 " 3 3(x - 3) In Exercises 31-35, find an ordered pair of real numbers that satisfies second of the given equations but not the first. 2-2 X s 5, tj — S s 5x 33. l^i = -^i, 2x - ij - 4 = x — 4 9 — 4 y ^ x-2 = 2 x-2 =JL -2 92 Separation and Related Concepts C-iarl-r „' ■ In Exercises 36-45, find the solution set of the given equation. 36. 3x - 4* = 8 - 5 37. % - 2) + 7{x - 3) = 30 42, M — £ a 2 x— 1 38. x + - + 1 = — x x 39. ^ = x x 40. x - 1 = x - 2 41. 2* - 2 = 2(x - 1) 43. * ~ * + «- 45 + ^4=3 x-1 x-2 x-3 44. *~? = 5 x-2 x-2 =2 / x-2 \ * x-3 x-3 \x-3^ CHAPTER SUMMARY There were three BETWKENNESS POSTULATES and three SEP- A RATI ON POSTULATES in this chapter. We list them below by name only. Try to state each of them in your own words. Draw a picture illustrat- ing what each postulate says. 9, THE A-B-C BETWEENNESS POSTULATE, 10. THE THREE-POINT BETWEENNESS POSTULATE. 11. THE LINE-BUILDING POSTULATE, 12. THE I-TXE SEPARATION POSTULATE. 13. THE PLANE SEPARATION POSTULATE. 14. THE SPACE SEPARATION POSTULATE. The following concepts were defined in this chapter. Be sure that you know all of them. SEGMENT INTERIOR OF A SEGMENT HAY INTERIOR OF A RAY OPPOSITE RATS ANCLE INTERIOR OF AN ANGLE EXTERIOR OF AN ANGLE TRIANGLE INTERIOR OF A TRIANGLE EXTERIOR OF A TRIANGLE QUADRILATERAL TWO SIDES OF A POINT ON A LINE TWO SIDES OF A LINE IN A PLANE TWO SIDES OF A PLANE IN SPACE HALFLINE OPPOSITE HALFUNES HALFPLANE OPPOSITE 1IALFPLANES ILYLFSPACK OPPOSITE H ALFSPACES Review Exerciset 93 A set of points is called CONVEX if for every two points P and Q in the set, the entire segment ¥Q is in the set. The null set and every set of points that contains just one point are also said to be convex. Each of the following sets is a convex set: segment, ray, line, plane, h&lflinc, halfplane, balfspaee, interior of an angle, and interior of a triangle. Six theorems were stated and proved in this chapter. Study them again so that you know and understand what they mean. The last part of this chapter contains a review of elementary algebra. You should know and be able to use the properties of equality and number operations that arc useful in solving equations. REVIEW EXERCISES In Exercises 1-15, indicate whether the statement Is true or false. L If points A, B, C and line t arc in the same plane, and if A and B are on the opposite sides of J, then C" must be either on the A -side of I or on the B-side of I 2. If B and C are two distinct points on the same side of line n in plane a, then every point of EC is on the B-side of n. 3. UA-B-C (point B is between points A and C) and B-D-Q then A-D-C and A-B-D. 4. If R-S-T and Q-S-T, then R-Q-S and R-Q-T. 5. If points R and S are on opposite sides of line m in plane a and points R and T are on opposite sides of m, then S and T are on the same side of m. 6. The betweenness relations R-S-T and R-U-T uniquely determine the order of the points R, S, T t U on a line, 7. If two rays intersect, they have one and only one point in common. 8. AJ9 = M. 9.AB = bX 10. opp A~S = BA 1L AB = BA 12. ab n bX = AB 13. AB U B~X = A~B 14. opp AB D opp B~A = A 9 15. If point B is between points A and C, then BA and BC are opposite rays. 94 Separation and Related Concepts Chapter 2 ■ Exercises 16-20 refer to Figure 2-4S. A 16. How many angles are determined by the segments shown In the figure? 17. How many triangles are determined by the segments shown in the fig- ure? Name them. 18. Point E is in the interior of two angles. Name them. 19. Only one of the labeled points in the figure is in the interior of each of three different angles. Name the point and name the three angles. 20. Two of Ae labeled points in the figure are not in tlie interior of any of the angles. Name them. 21. Draw A ABC. Mark a point D such that D is between A and B. Mark a point E such that B is between C and E. (a) Docs ED intersect AS? Why? (b) Does EZ> intersect BC? Why? (c) Does ED intersect A£? Why? 22. Define the interior of £PQR> 23. Cicen: Line 1 f£ line m. Point C is between points A and B oa liue L Point C is between paints D and E on line m. Point C is between points C and E. Point F is between points A and E. (a) Which point (C, G, or F } is In the interior of I ACE? (b) Indicate whether each of the following is true or false. (1) Point F is on line I, (2) Point F is on line iiu (3) Point F is a point of / ACE. (4) Point C is a point of L BCF.. (5) Line FG does not intersect AC. Review Exercise* 95 24. If r f $, t, u are four distinct coplanar rays having a common endpoint and if no two of the rays arc collinear, how many angles are formed by these rays? 25. Which of our postulates guarantees that a halfplane is a convex set? 26. Is the intersection of two convex sets always a convex set? 27. Explain why the interior of an angle is a convex set. 28. Is a line with one point deleted a convex set? 29. Is the. union of two convex sets always a convex set? 30. Describe two convex sets whose union is a convex set. In Exercises 31-35, solve for x. 31. 3x - 5 = Ix - 25 32. 2(r - 5} ■ 3x - 5 33.^1+1=^ 34, 1.75x = 17.50 35. x - 1 + 2(x - 1) = 17{* - 1) + 14 In Exercises 36 --40, find the solution set 36, x + x = 2* 37. x + 1 = x + 2 38. X 39. x-1 2*-2 _ 1 " 2 40, X - 1 = Iff* -9 Chapter HeUa Hammld/Bopho GuiUumetttt Distance and Coordinate Systems 3.1 INTRODUCTION At this point in our formal geometry we have do postulates con- cerning the sizes of objects. We have no basis for saying how big an object is or even for saying that one object is bigger than another. Al- though the word "size" is often used in informal speech* it is not in the official vocabulary of formal geometry. Instead we talk about the length of a segment, the measure of an angle, the area of a rectangle, the volume of a sphere, the distance between two points, and so on. In elementary mathematics it is customary to draw a horizontal number line with numbers increasing from left to right and a vertical number line with numbers increasing in the upward direction. In some applications it may be more convenient to order the numbers from right to left or from up to down. A number line, such as the one in Fig- ure 3-1 in which the numbers increase from left to right, is a good de- vice for illustrating certain relationships among real numbers. One of them is the order relation. The fact that 3 is greater than 1 is consistent With the fact that 3 is to the right of 1 in Figure 3-1. The fact that 1 is to the right of —2 agrees with the fact that 1 is greater than —2. ! ! ► -4-3-2-1 I 2 3 4 5 6 Figure 3-1 93 Distance and Coordinate Systems Chapter 3 Another relationship is based on the betweenness relation for real numbers. Since — is between — ■ and %, the point marked — lies 3 5 § 3 between the points marked — and — on the number line shown in Fig* ure 3-2. Figure 3-2 This chapter is about distance and coordinate systems. We begin by considering the distance between two points. The idea of a coordinate system on a line is an extension of our ideas about a number line. Co- ordinate systems are useful in developing the properties of distance. 3.2 DISTANCE Asking "How long is a certain segment?" is equivalent to asking, "How far apart are the endpoints of that segment?" In the world of real objects we can answer the question, "TTow far apart?" by using a physical ruler. We might determine that two points P and Q are 2 yd. apart, or 6 ft. apart, or 72 in. apart To make a physical ruler graduated, say, in inches, we must know what 1 im is. We must have a segment 1 in. long. In the United Stales the accepted relation between inches and meters is 39.37 in. = 1 meter. For many years the meter was described officially by two marks on a phttinum-iridium bar kept in France. These marks repre- sented the eildpointS of the segment that was the official meter. Al- though the modern standard for measuring distances is now based on the wavelength of a certain kind of light, the idea that a given segment may be a standard or unit for measuring distance is important in our geometry. Tn the remainder of this section we discuss informally some of the basic properties of distance and then state these properties for- mally as postulates. Given any segment, say FQ y we could agree that this segment is the unit of distance. Thus we might start with a given stick and say, "Let the distance from one end of this stick to the other end be 1, and let us call this unit of distance the "stick/' Then the given stick or unit is the basis of a system of distances which we might call the stick-system. In this system the distance between any two different points is a posi- tive number. In particular, the distance between the endpoints of the 3.2 Distance 99 unit stick is 1, Tf the distance between points R and S is 3 in this system, this means that R and S are 3 times as far apart as the endpoints of the unit stick. If the distance between points U and Vis 1 in this system, this means that V and Vare just as far apart as the endpoints of the unit slick. Speaking more formally, each segment PQ determines a distance function. The domain of this function is the set of all segments, or if you prefer, the set of all pah's of distinct points. The range of this func- tion is a set of positive numbers. It is convenient in developing the con- cept of distance to take the distance between a point and itself to be 0. Then the domain of a distance function is the set of all pairs (not neces- sarily distinct) of points and the range is a set of noimcgative numbers including 0. Later, after we adopt the Ruler Postulate, it will be obvious that the range is the set of all nonnegative numbers including 0. Our first Distance Postulate is related to the idea that any stick can serve as a unit of distance. We mi^it develop a formal geometry with one distance f miction. This would be like using inches for all distances whether they are thicknesses of paper or distances between stars. Our first postulate, the Distance Existence Postulate, reveals our preference far recognizing the possibility of various units of distance i n our formal geometry. The other Distance Postulates are motivated also by phys- ical experiences with distance, betweenness, and separation. If A, B, C are three distinct eollincai points with 8 between A and C, then wc want the distance between A and B plus the distance be- tween B and C to be equal to the distance between A and C as illus- trated in Figure 3-3. 14* a ' T c Figure 3-3 If A, B, C are three noncollinear points, then we want the distance from A to B (the same as the distance between A and B) to be less than the distance from A to C plus the distance from C to B as illustrated in Figure 3-4. Figure 3-4 100 Distance and Coordinate Systems Chapter 3 We are now ready to State our Basic Distance Postulates in winch we exercise extreme care in the choice of words so that we agree on exactly what they mean. As always we need to know precisely what we are accepting without proof in the building of our formal geometry. We use "unique" to mean "one and only one" or "exactly one." Basic Distance Postulates POSTULATE 15 (Distance Existence Postulate) If AB is any segment, there is a correspondence which matches with every segment €D in space a unique positive number, the number matched with AJi being 1. If C and D are distinct points, there is a unique positive number matched with CD according to Postulate ] 5, We may also think of this Dumber as matched with the set (C, D). If we associate this number with the segment CD, we think of it as a length. If we associate the number with the set {C, 1?}, we think of it as a distance. Postulate 15, however, says nothing about the distance from C to D if C = D; in other words, it says nothing about the distance between a point and itself. We take care of this with a definition. Definition 3.1 The distance between anv point and itself is a Definition 3,2 L The correspondence that matches a unique positive num- ber with each pair of distinct points C and D t as in Postu- late 15, and the number with the points C and D if C = D. as in Definition 3. 1 , is called the distance function determined by AB or the distance function based on JT& 2. The segment AB that determines a distance function is the unit segment for that distance function. 3. The number matched with C and D, as in Postulate 15, is the distance from C to D or the distance between C and D or the length of CD. Notation. If P and Q are any points, not necessarily distinct, then the distance between P and Q in the distance function based on AB is denoted by FQ{in AB units) or simply by FQ if the unit is understood. 3.2 Distance 101 Example f (Informal) Suppose AB is a segment 1 in. long, CD is a segment 1 ft. long, and EFis a segment & in. long. Then EF (in AB units) = 6 and EF (in W unite) = |. Although Postulates 16 and 17 could be omitted and proved as the- orems after the Ruler Postulate is adopted, they are included in our formal geometry in order to simplify the development. POSTULATE 16 (Distance Betweenmss Postulate) Tf A, B, C are collinear points such that A-B-C, Lhen for any distance function we have AB + BC = AC. (See Figure 3-5.) a B c 4 • • m ► Figure 3-5 Example 2 If A-B-C, AB - 6, BC = 8 P then AC - 14. POSTULATE 17 (Triangfe Inequality Postulate) If A, B, C are noncollinear points, then for distances in any system we have AB 4- BC > AC (Sec Figure 3-6.) Figured Example 3 If A, B, C are noncollinear points and if BC = 10, then BA + AC> 10. Example 4 If the lengths of the sides of a triangle are x, y t z T then x + y > %, y + z>x, and z+ x> y. Suppose that FQ is a segment 1 in. long and that 1£S is a segment 1 ft. long. Suppose that A and B are two different points and that C and 102 Distance and Coordinate System'. Chapter 3 D are two different points. We use distances to compare how far A is from B with how far C is from D, If we say that it is 3 times as far from A to B as from C to D, we mean that AB - 3 CD, or that ^ = 3. ^ CD (See Figure 3-7.) Our experience with physical measurements tells us that we should get the same comparison regardless of which distance function we use. Thus .r AB (in PQ units) n . AB (in M units) rt , * W (in TQ units) = 3 ' *" CD (in M units) = 3 ^ For example, if AB = 72 (in inches) and CD = 24 (in inches), then AB = 6 (in feet) and CD = 2 (in feet). Thus AB (in inches) _ 72 _ AB (in feet) 6 CD (in inches) "" 24 " CD (in feet) " 2 = This is the basis for our next postulate. POSTULATE 18 (Dhtance Ratio Postulate) If PQ and RS are unit segments and A, B t C, D axe points such that A^B.C^A then AB (in 2^ units) AB (in M imits) GO (in PQ units) "" CD (in 15 units) or, equivalently, AB (in /p units) CD (in PQ units) AB (in RS units) " " CD (in R5 units) " In making physical measurements we recognize that there are many accurate foot rulers. Measurements made with an official foot ruler and an accurate copy of one should agree. Let us see how Postu- late 18 is concerned with this. Suppose that PQ and H5 are unit segments and RS (in FQ units) = 1. This means that the length of RS in PQ units is 1 or, infor- mally, that M is a copy of FQ. If A and B are any points, it follows from Postulate 18 that AB (in TQ waits) __ AB (in A S units) RS (in PQ units) " RS (in M units) " 3.2 Distance 103 Since we assumed that RS (in FQ units) = 1 and since RS (in Jf3 units) = 1 by the Distance Existence Postulate, it follows that AB (in TQ units) AB (in RS units) 1 1 or AB (in TQ units) = AB (in RS units). This proves the following theorem. THEOREM 3- 1 If FQ and RE are segments such that the length of -RS in Pp units is 1, then for all points A and B it is true that AB (in M units) = AB (in PQ units). KXERCISJiS 3*2 1. If A and B are points and if CD is a segment, which of the following are necessarily true about the number AH (in CD units)? (a) It is a real number. ;b) It is a positive number. (c) It is a nonnegaiive number. (d) It is an irrational number. 2. If you know that A and B are distinct points and that W> is a segment, what can you say about the number AB (in RS units)? 3. If you know that A-B-C, what can you say about the number AB + EC , AC 4. If you know that A, B, C are noncoHinear points, what can you say about the number AB + BC ? 5. If you know that A, B, C are points, that A # G* and that FQ and H5 are segments, what can you say about the numbers AB (in FQ units) , AB (in RS units) . AC (in FQ units) ! AC (in RS units) Why is it not necessary to say A ^ C in Exercise 4? 6. If you know that A, B, C are noncoUinear points and that PQ and H5 are segments, how does the number AB (in FQ units) + BC (in FQ units) \C (in FQ m.iiHi compare with the number AB (in H5 units) + BC (an BS units) a AC (in RS units) 104 Distance and Coordinate Systemi Chapters 7. Let A, B> C, D, E be distinct points ordered on a line I as shown in the figure, andjttjually spaced so that AB (in AB units) =: BC (in Tfil units) = CD (ia AB units)_= D£ (in ZB units). Find the following: AB (in BC units), £C (in BC units), CD (in BC units), and DE (in BC units). ABODE 4 . . . « . +t 8. Given the situation of Exercise 7, find the distances AB t BC, CD, DE, all In AC units, », Given the situation of Exercise 7, And the numbers AB (hi AB units), AB (in AC units), AB (is Bl? units), and AB (ii i i 7 ! i 1 nits). 10. Civen the situation of Exercise 7, find AD (in A"B units), AD (in BC units), AD (in AC units), AD (in BE units), and AD (in AE units). 1 1. Given the situation of Exercise 7, copy and complete the following proof that BD (in JU units) = 1. Proof: Expressing ail distances in ACT units we have the following: 1. BD — BC + CD 1. Distance Between ness Postulate CD 2. -jg = 1 2. Exercise 7; Distance Ratio Postulate 3. CD = AM 3, Multiplication Property of Equality 4. BD = BC + AB 4. Steps i, 3; Substitution 5. BD = AB + BC 5. a AC = AB + BC 6. Q] 7, BD = AC 7. & AC = 1 8. [7] 9. BD = 1 9. 12. In the Distance Ratio Postulate, take PQ —CD and M = JB. Using this special case of the postulate, prove that AB (in CD units) - CD (in AB units) = L (Does this result seem reasonable? Think of AB as a stick 2 ft. long and CD as a stick 3 ft. long. Then AB (in CD units) = |, CD (in AS units) = 1 1 Land*- 4 =10 13. Given A, B, C, D such that AB (in C~D units) = % find CD (in AB units). (Does your answer seem reasonable? The given information means that AB is — as long as CD. Your answer means that C~D is how many times as long as AB?) 14. challenge FfioBLEM. Given segments AB, CJ5» EF, prove that EF (in AB units) - EF (in CD units) • CD (in JF units). (A simple example is: The number of feet from E to F equals the num- ber of yards from E to F times the number of feet in a yard.) 3.3 Une Coordinate Systems 105 15. {An Infortnal Geometry Exercise.) fat the figure, suppose that the part of I between F and M is on the edge of a ruler. On this ruler is assigned to P and 1 to Q. What numbers should be assigned to C, D t E, and \f? p q CD KM 4 — * ir « — r 9—9 — M 16. (An Informal Geometry Exercise.) Two number lines / and m are placed parallel to each other as shown in the figure below. What num- bers should be assigned to S and T on IP What numbers should be as- signed to F, Q, and H on m? A s T -I 4 p —m — 2 Q •— V m 100 3 a 4 1 17. Refer to Exercise 16. If A on if is directly alxrve B on m and if 100 is as- signed to A, what numl>er is assigned to B? 18. challenge problem. Hefer to Exercise 1 6. If a number x is assigned to a point on land if the number x' is assigned to the point directly be- low it on m f express x' in terms of x. 19* In the Distance Hah'o Postulate we stated that the two equations were equivalent, (a) Derive the second equation from the first one using some of the properties of equality, (b) Derive the first equation from the sec- ond one using some of the properties of equality. 3.3 LINE COORDINATE SYSTEMS A key feature of a number line is that different points are matched with different numbers. In fact, we consider al the points of the line as matched with all the real numbers. Such a matching is called a one-to- one correspondence between the set of all points on the line and the set of all real numbers. Another key feature of a number line is that the distance between any two points is the absolute value of the difference of the numbers matched with those two points. A unit segment for these distances is the segment whose endpoints arc matched with and 1. Example '" Figure 3-8, AB= 1-1 -(-2)| = 1. J BC=|0-(-l)| = |l| = l. CE = |2 - 0| = 2. AC = |0 - (-2)| = [2| = 2. BD = |1 - (-1)| = 2. BE = 2 - (-1)| = |3| = 3. AE= |2 - (-2)1, = 4. AE = |2 - <-2)| = |4| = 4, A B C D S 106 Distance and Coordinate Systems Chapter 3 The concept in formal geometry that corresponds to a number line in informal geometry is the concept of a coordinate system on a line. Coordinate systems are useful tools in our formal development of ge- ometry for planes and spaces as well as for lines. Definition 3,3 Let FQ be a unit segment and / a line. A coordinate system on I relative to PQ is a one-to-one cor- respondence between the set of all points of I and the set of all real numbers such that if points A. B, Care matched with the real numbers a* b t c, respectively, then 1. B is between A and C if and only if b is between a and c and 2, AB (in TQ units) = \a - b\. Definition 3,4 2* The origin of a coordinate system on a line is the point matched with 0, 2. The unit point is the point matched with 1. 3. The number matched with a point is its coordinate. These definitions describe and help us to think about a coordinate system. But a very important issue must not be overlooked here. How do we know there are things such as coordinate systems in our formal geometry? To answer this question we return to our experiences with number lines and physical measurements. Let us suppose that TQisa. unit segment, say a segment 1 cm, long, and that A and B are any two distinct points on I. From our experience with physical rulers we know that we can lay a ruler graduated in centi- meters alongside I and measure distances starring from A and extend- ing toward B as shown in Figure 3-9. ♦I Ffaove 3-D l' entire cere Of course, we can just as well start at B and measure toward A as shown in Figure 3-10, These ideas suggest our next postulate. -*/ Figure 3-10 Centimeters 3,4 Rays, Segments, and Coordinate* 107 POSTULATE 19 (Ruler Postulate) If AB is a unit segment and if P and Q are distinct points on a line l t then there is a unique coordi- nate system on / relative to AB such that the origin is P and the coordi- nate q of Q is ft positive number. (See Figure 3-11.) I q(q>0} Figure 3-11 We are now ready for a theorem. THEOREM 3,2 (The Origin and Unit Point Theorem) If P and Q are any two distinct points, then there is a unique coordinate system on PQ with P as origin and Q as unit point. Proof: 1. There is a unique coordinate system 1. Ruler Postulate on PQ relative to TQ t with the coor- dinate of P equal to and the coor- dinate q of Q a positive number. 2. PQ (in Pp units) - q - = q. 2, [T] 3. PQ {in Pt> units) si 3. [7] 4 q = 2 4 5. Q is the unit point. 5, [7] 3,4 RAYS, SEGMENTS, AND COORDINATES I^t A and B be two distinct points on a line I. We know from the Origin and Unit Point Theorem that there is a coordinate system $ with A as origin and R as unit point. Let X be any point of I and let x be its coordinate in %, Then it follows from the definition of a coordinate system that A is between X and B if and only if is between x and lj X is between A and jB if and only if x is between and I ; B is between A and X if and only if 1 is between and x. (See Figure 3-12,) X A a X A X s 1 A JC 1 B X I X Figure 3-12 108 Distance and Coordinate Systems Chapter 3 Now is between x and 1 if and only if t < 0; x is between and 1 if and only if < % < 1; and ! is between and x if and only if .* > 1 . It follows from the definitions of ray, ray opposite to a rav, and segment that: ojtpAB = {X :x<0) 4 2 A i >t AE = {X : < x < 1} 4 ^ — G- A$ = {X : x > 0} 4 4 — $- -►/ 1 = {X i x < I ) < d — £ +f ■w Similarly, if C and D are two distinct points on a line I with coordi- nates 2 and 5, respectively, and if X is a (variable) point on / with co- ordinate x f then: <5B = {X : 2 < x < 5} a {X : i > 2} = {X:x< 5} opp CD a (JE : x £4} opp D? = {X : i > 5} CD = {X : .r is real} .' £_ % 2 5 rl _s 9 « — ' 2 5 w 1 6 V 2 3 tl 4 — —A 23 2 5 C 9 — • , 2 5 r t £ Q * i More generally, if the coordinates of two distinct points A and B on a line I are a and b, respectively, then it is convenient to consider two cases in expressing subsets of AB using set-builder symbols as follows. If <i < b If a > fr AB = {X;a<x<b) IB = {X : b < % < a) AB= {X:x> a) AB = {X : x < a] BX = {X : x < b) ~BA = [X:x>b] oppA$={X:x<a) oppA& = {X : x > a} oppBA = (X :x>b] oppBA={Xix<b) 3.4 Rays, Segment*, and Coordinates 109 EXERCISES 3.4 In Exercises J -5, a line / with a coordinate system $ is given. The coordi- nates of points A,B t C arc — 3 t 0, 4, respectively. In each exercise draw the graph of the set and express it in terms of coordinates using a set-builder symbol Exercise I has been completed as i sample. 1. BC Solutions BC = {X : x > 0} 4 fc -3 + 1 2. BA 3. AC 4. m 5. £2 In Exercises 6-10, a line / and a coordinate system § are given. In each exercise, given the coordinates of two points, find the coordinate of a third point Exercise 6 has been worked as a sample. {Note: cd A means "coordi- nate of A.'*) a cd A = 2, cd B = 5, cd X = x. If X £ ~&& and AX = 2 ■ AS, find x. Solution x < 2, AX = 2 - x, AB = 3. % - x = 6, x = -4 I i ft. -N 7. cd C = U cd D = 1, cd P = p. If P € c3 and ■—- = ^, find p. 8. cdE = -5, cd F = G,cdQ = qAfQ£ oppFE and EF = FQ, findq. 9. cd C = 29, cd H = 129, cd I = i. If GI = III, find L 10, cd J = 15. cd K = Q> cd R = rAf JR = 2- RK T find the two possible values of r. In Exercises 11-15, given the coordinates of two points on a line, find the length of the segment joining these two points, U. cd A m 5, cd B = 173 12. cdC= ~5,cdB= 173 13. cdC = -& cdD = - 173 14. cd E = 147,5, cd F = 237.6 15. cd G = 19j, cd f* = - 17| 1X0 Dictince and Coordinate Systems Chapter 3 ■ In Exercises 16-20, given the coordinates of two points on a line, find the coordinate p of a third point P satisfying the stated condition. 16. cd A = 5, cd B = 10, P € Ail and AP = 5 17. cdA = o,cdB = 10, P € oppBA and AP = 5 18. «f A = 5. erf B = 10, P € BA and AP = 5 Id. erf A = 5, erf B = 10, P € opn A^ and AP = 5 20. cdC= -10, cdD = Q, P£ CD and PC = 17| ■ In Exercises 21-25, given a line I and a coordinate system on it in which x is the coordinate of a variable point X, draw a sketch showing / and the given subset of/. 21. {X : ~6<x< -2} 22. {X :x< -6} 23. {X : x > -2} 24. {Xix< -6orz>; -2} 25. {X\x< 10} ■ In Exercises 26-35, given a tine with points and coordinates as marked in Figure 3-13, find the given distance. CQ D E F Q H P / JS X L MT N -3 t -2 -I 1 2 f 3 4^6 G 7*8 3 V2 6 /£ -V fl Figure 343 26. CE 31. /\) 27. DC 32. st 28. HP 33. SF 29. IC 34. A/g 30, FQ 3d. PS L0 In Exercises 36-45, given a line and a coordinate system in which cJ A = 12, erf B = - 8 f cd C = 0, erf D = -4.2, and erf E = \M deter- mine if the given between ijcss relation is true or if it is false. 36. A-B-C 41. B-E-A 37. A-B-D 42. D-C-A 38. D-C-E 43. D-B-A 39. C-J5-A 44 B-E-D 40. £-C-B 45. C-A-B 3-5 Segment! and Ccmgrusnc* 111 In Exercises 48-50, given that AS is a segment I ft- long and that CD is a segment 1 yd long, find the given distance. 46. EF (in AS units), if EF (in CD unite) = 5 47. CH (in W units), if GH (in AB units) = 5 48. CD (in CD units) 49. CD {in Al units) 50. AB (in (715 units) 51. Copy and complete the proof of Theorem 3.2. 3.5 SEGMENTS AND CONGRUENCE In Chapter 5 we develop in considerable detail an idea called the congruence idea. Informally speaking, two figures arc congruent if they have the "same size and shape." The terms "size** and "shape" are not considered as a part of our formal geometry. In elementary geometry, we develop the concept of congruence for segments, for angles, and for triangles. The concept of congruence for segments is easy and is appropriate to include here. Intuitively, we feel that all seg- ments have the same shape; hence, they have the same size and shape if they have the same length. We make this formal in the following definition. Definition 3.5 Two segments (distinct or not) are congru- ent if and only if they have the same length. If two segments are congruent, we say that each of them is congruent to the other one and wc refer to them as congruent segments. It is convenient to have a special symbol for congruence; thus AB s CD means AB is congruent to (35, It may be helpful to compare the words "congruent and congru- ence" with the words "equal and equality." 7=3 + 4 may he read as "7 is equal to 3 plus 4" 7 — 3 -f 4 is an example of an equality. ABs UD may be read as "AB is congruent to (?D" AB as CD is an example of a congruence, When working with segments it is important to note carefully the difference between equality and congruence. The statement AB = CD means that A"B and UD are the same set of points, that is, that "AT?" 112 Distance end Coordinate Systems Chapter 3 and "CD" are different names for the same segment. This statement is true if and only if A = C and B = D, or A = D and B = C. (See Figure 3-14.) • • Or • • C D D C Figure 3-14 The statement AB at UD means that AB and CD have the same length, that is, AB = CD, Therefore, if AB at W f then AB = CD; and if AB = CD, then_ AB at CD. Notethat AB = CD implies 33 « CD but that XB at CD does not imply AR = CD. We have emphasized the difference between congruence and equality as these concepts apply to segments. We note next some similarities. Recall the equivalence properties of equality, the reflexive, symmetric, and transitive properties. (See Section 2.8.) Congruence for segments has the same properties, as stated in the following theorem. THEOREM 3.3 Congruence for segments is reflexive, symmetric , and transitive. Proof; Let AB be any segment Then its length AB is a number, and AB = AB by the reflexive property of equality. But AB = AB implies that ~KB s AR Therefore congruence for segments is reflexive. (The remainder of this proof is assigned as an exercise.) Our next theorem expresses formally a simple idea concerned with adding or subtracting lengths. The theorem expresses this idea formally in terms of congruences. Theorem 3.4 is followed by Corollary 3,4.1. A corollary i$ a theorem associated with another theorem from which it follows rather easily. THEOREM &4 (The Length-Addition Theorem for Segments) If distinct points B and C are between points A and D and if A3 as CD, then A€ss 3D. Proof: There arc two possibilities as suggested in Figure 3-15. i— g g g a SB a Figure 3-15 3.5 Segmmts and Cortgru«rtoe 113 Case L If A-B-C, then R-C-D, and it follows from the Distance Be- twecnness Postulate that AB + BC = AC and BC + CD = BD, Since AS = CD (hypothesis) and since BC = BC (Why?), it follows from the addition property of equality that But + BC=CD + BC. CD + BC = BC + CD, (Why?) Therefore AB + BC= BC+ CD (Why?) and AC = BD. This proves that AC ss ED, Case 2, If A -G-B, then C-B-D, and it follows from the Distance Be- tweenness Postulate that AC + Cfl = AB and CB + BD = CD. (The remainder of this proof is assigned as an exercise. Note in Case I that the idea, or strategy, of the proof is the addition of lengths. The proof of Case 2 involves the subtraction of lengths.) COROLLARY 3,4,1 If distinct points B and C are between points A and D and if AC == BD, then AB =ss £J5. /too/* If B and C are between A and D, then C and B are between A and P. The corollary follows immediately from Theorem 3.4 by inter- changing B and C that is, by renaming point B as point C and renam- ing point C as point B, COROLLARY 3.45 If A, B, C, D, £, £_are points such that A-jB-C D-E-b\ AB =s 73E, BC ssEF, then AC s 35F. iVoo/- Assigned as an exercise. COROLLARY 3A.3 If A, B s C, D, E t F are points such that A-B-C, Z>-£-i? AB == 0E, A^s Z5F, then BC - £F. Bkx>/; Assigned as an exercise, 114 Distance a raJ Coordinate Systems Chapter 3 We now raise a question that leads to the final theorem of ibis sec- tion. Suppose that a distance function determined by PQ, say, and a ray AB are given. Is there a point C on AB such that AC = 7? Is there only one such point? We know from the Ruler Postulate thai there is a unique coordinate system on AB relative to FQ such lhat the origin is A and the coordinate b of B is a positive number. Then a3= {X:x>0). (See Figure 346.) J L b * Figure S-IB This means that there is a one-to-one correspondence between the set of all points of AB and the set of all nonncgative numbers. There- fore there is exactly one point C on AB such that Lhe coordinate of C is Z Then AC = 7 - = 7. If D is any other point of AB, then the coordinate d of D is not 7 and AD = d - ^ 7, Therefore there is one and only one point C on AB such diat AC = 7. We know from our experience with rulers that we can start at any point on a line and 'lay off' in either direction (along either of the two rays on the line that have the starting point as its endpoint) a segment of any desired length. The example we discussed shows that in our for- mal geometry, given any ray, we can 'lay off," or "construct," a seg- ment which Is 7 units long such that the endpoint of the my is one of die endpoints of the segment. Of course, we could start with any posi- tive number other than 7 and the same reasoning would apply. We could start also with a given segment and "lay oF* on AB a segment whose length is the length of the given segment, thai is, a segment con- gruent to the given segment. The "lay ofF" or "construct" language is informal All we are really saying is that there is such a segment and that it is unique* These ideas lead to our next theorem. THEOREM 3.5 (Segment Construction Theorem) Given a seg- ment CD and a ray AB, there is exactly one point P on AB such lhat APssCD. 3.5 Segment* and Congruence 115 Proof: Given a ray A3, a segment CD with CD = p t and a distance function » there is a unique coordinate system § on AB such that A is the origin and such that the coordinate h of B is a positive number. (See Figure 3-17.) Then "*= iX:x>Q\, 4 =_£ b P Figure 3-17 Let P be the unique point of AB such that the coordinate of P is p. Then AP = p — = p. If Q is any point of AS other than P t then its coor- dinate q is different from p and = <j-0=q=£p. Therefore there is only one P on A3 such that AP = p, that is, such that EXERCISES 3,5 In Exercises I~10 t a line I with a coordinate system $ is given. The coordi- nates of points A, B, C, D are —5, 3, 5 t 13, respectively. (Sec Figure 3-18.) In each exercise, determine whether the given statement is true or false. A B C D -5 3 5 * r 13 figure 3-1 I. "KB ss CD 6. AB=sB(7 I.BCs^W 7. AB= lA 3. ATszW 8, JB= £D 4. AB + BC = AC 9. AC nlE5 = riJC ,iASUl? = JTC 10. AC + HD = : AB + CD + 2-BC Iii Exercises 1 1—1-5 name the property of congruence for segments which justifies the given statement. 1L If AB at UD t then ITS a ATI 12. If MssUDimdrBsiDE* thenAEsDE 13. FQ as £P 14. If^^^BU^CD ( andOT-EF. thenAB^EF. 15. If XY is a segment, then XT S£ XY. ilfi Distance and Coordfnate System t Chapter 3 16, Complete the proof of Theorem 3.3, 17. Complete the proof of Theorem 3.4. IS. Draw an appropriate figure and prove Corollary 3,4.2. 10. Draw an appropriate figure and prove Corollary 3.4.3. ■ In Exercises 20-25, the coordinates of points B and C in a coordinate sys- — ■ tern on HC are given. If A-B-C and ~KE ss BC, find the coordinate of A. 20. cdB = 0,cdC=5 23, cdB = $, cdC = $ 21. cd B=5,cdC= 24. cd B = 159, cd C = -156 22. cd B= ^234, cd C = - 108 25, cd B = 27, cd C = 102 3.6 TWO COORDINATE SYSTEMS ON A LINE Let A and B be the origin and the unit point, respectively, of a co- ordinate system § on a line /, What is the coordinate of the point P if P is between A and B and two-thirds of the way from A to B? (See Figure 3-19.) Obviously, adf = | since AP = § and AS = 1. * • ■ * M Figure 3-19 I Let C and D be points with coordinates 35.0 and 39,8, respectively, on a line /. What is the coordinate q of point Q if it is between C and D and two-thirds of the way from C to D? (See Figure 3-20.) G q D 4 — » + 9—tt 35.0 ? 39-B Figure 3-20 * 1 We know that there is another coordinate system on / in which the coordinate of C is and the coordinate of D is 1. If Q is thought of as a variable point on this line with q as its coordinate in one system and % as its coordinate in the other system, then the particular point Q we want lias j as its s-coordinate. In this section, we learn how to express the ^-coordinates in terms of the ^-coordinates. In this example, the re- lation is q = 35.0 + 4.8*. Substituting § for x we get q = 35.0 + (4.8)(|) = 35.0 + 3.2 = 38.2, Therefore the point that is two-thirds of the way from the point with coordinate 35.0 to the point with coordinate 39.8 is the point with co- ordinate 38,2. 3.6 Two Coordinate Systems on a Line 117 The relationship between two coordinate systems on a line is useful in solving exercises involving points of division like the "two-thirds of die way from C to D" exercise as well as some exercises appearing later in this chapter. This relationship is also useful in gaining new algebraic insights. In studying equations like 1/ — 2 -f 3x and Jt- 2 1 it is sometimes helpful to think of them geometrically in terms of co- ordinates on a line. In later chapters Wfl consider coordinate systems in a plane and in space. In studying a line in a plane or in space you will find it helpful to think of several coordinate systems associated with the line. What we do later is a natural extension of the groundwork we are laying in this chapter. We begin with a lemma, a "little theorem," that is useful in proving a "big theorem." Lemma 3.6. J plays a key role in the proof of Theorem 3.6, which is followed by Corollary 3.6, .1. LEMMA 3.6.1 Let Xi and x 2 be the coordinates of distinct points Xi and AV respectively, on a line /. If x is the coordinate of a point X on /, then AXi X — X\ if X<= XiX 2 A 2 A a % 2 — Xi and if X £ opp A'jAV Proof: First, suppose that X £ XiXi; then either xi < x;? or x 3 < *i- (See Figure 3-21.) If x t < xg, then xi < x, XX a = x - x u XgXj = *2 — *i, and AXi x — Xi X2X-1 X 2 — Xl XX, X - *1 X2X1 X2 — x L ■> X%Xi X2 — x% Xi Xz X X X 3 Xi • • • M 4 • • • hi X\ < 12 JC2 < X\ Figure 3-21 If *a < xi, then x < xi, XXi = x\ - x t X 2 Xi = Xi - x 2 » and XX 1 X\ — X X — Xi X2X1 l! - X 2 X 2 - Xi 118 Distance and Coordinate QyitNM Chapter 3 Next suppose that X £ opp XiX 2 ; then either x x < x 2 or x 2 < x\. (See Figure 3-22.) If Xi < x 2j then t\ > x t XX\ = X] — x, X 2 Xi s %2 — Xu ^d \ A: 3Ci — x x — *i X 2 Xi *2 — *i % — a& ' * • • -• ► / 4 • • • ►i x 11 ar 2 a- 3 *! j x 1 < *a *a < *i Figure 3-22 If xa < xi, then r > x lt XXi = x — x 1? X 2 X t = x% — x 2 , and XXi x — Xj x — Xi X 2 Xi Xi - Xa x 2 — X! ' Therefore, if X £ SjXj, then XXi x- Xi XgXj X2 — X] regardless of whether Xi < *2 or x T > x 2 . Also, if X € opp XiX 2 , then XX 1 x — xi XjjXi x 2 — Xi regardless of whether x-i < Xa or xi >• X2. THEOREM 3.6 (The Two Coordinate Systems Theorem) If Xx and X 2 are two distinct points of a line I, if the coordinates of Xi and X 2 are *i and x 2 , respectively, in a coordinate system S, andx^ and x 2i respectively, in a coordinate system §*, then for every point X on /, it is true that x — %\ x / — xj Xt — Xi x 2 — x t where % and x' are the coordinates of X in $ and in §', respectively. Proof: Suppose that we are given a line / and points X x and X 2 on I with coordinates Xi and x 2 in § and coordinates xi and x 2 in §' as in the statement of the theorem. (Sec Figure 3-23.) * • • 7ZI ►* Figure 3-23 x\ x' t {,$'} 3.6 Two Coordinate Systems on a Line 119 Suppose that X" £ X1X2; then it follows from Lemma 3.6.1 that XXj x- xi XXi xf - x^ x~ ^1 x 1 - xj X2Xi x 2 — xi ' X^Xi x 2 — x{ * x 2 — xi x^ — x[ *2 — X? — *J *t — *f x 1 «J Suppose that X £ opp XiX 2 ; then it follows from Lemma 3.6. 1 that XXj *- *i X 2 X t XXi x — Xi *2 — *1 X' 2 — X[ Therefore for every X on I it is true that x — x\ x 1 — x[ X'i — X\ x 2 — x\ and the proof is complete. Theorem 3.6 is closely related to the Distance Ratio Postulate. Ac- V X cording to this postulate the ratio M s independent of the distance XzXi function. Tf we use the distance ftmction of the coordinate system g in Theorem 3.6 y we have XX j |x — X\\ X2X1 \*2 - Xi\ If we use the distance function of the coordinate system §' in Theorem 3,6, we have XXi = K - «;i x 2 x, 14-^1' Therefore it follows directly from the Distance Ratio Postulate that I x — Xi I I x? — xj I I Ig — *1 I I * 2 — X[ \ It follows from the Two Coordinate Systems Theorem that the equa- tion obtained from this one by omitting the absolute value symbols is also true. 120 Distance and Coordinate Systems Chapter 3 COROLLARY 3. 6.1 Let Xi and X 2 be the origin and unit point, respectively, in a coordinate system §»&. on a line L Let xi and % be the coordinates of X t and X 2j respectively, in a coordinate system § r on /. Let k and x be the coordinates of a point X on / in the sys- tems %k and $ x , respectively. Then (i) and W x — X\ X2 — Xi a k t that is, x = xi + k(%2 — *i)- Proo/' (See Figure 3-24.) It follows from Lemma 3.6.1 that XX, k- or But XbX l 1 - XXi k-Q X2X1 = I - = * s -*. XX XjX L > a Why? Therefore and (!) is proved. K%Ki 1 *' x a x 1 1 (£*) r *1 X2 * CgJ Figure 3-24 It follows from Theorem 3-6 that x - - xi k - -0 Xi - xi 1 — 3,6 Two Coordinate Systems on a Line 121 Hence it follows that X2 — Xi X = *i + k(x 2 - X& and so (2) is proved. Example / (See Figure 3-25.) Think of A as X h B as X 2 . Then Xi = h X2 = 7, Xj = 3, *2 = 15, Then it follows from the Two Coordinate Systems Theorem that x- 1 x*- 3 7-1 15-3 for every point X on AB. This equation may be solved for x in terms of X* or for x* in terms of x. The resulting equations are S? ss §1? — 4 and x* = 2x + 1. * • • ♦ * S 15 ^ (S'j Figure 3-35 These equations are useful in finding r' if you know x or in finding x if you know x\ If you are given that x' = — 6, you can use the first of these equations to get x = — 3j. If you are given x = 2, yon can use the second of these equations to getx' = 5, Example 2 (See Figure 3-20.) Think of P as X L , Q as X 2 . Then ar*-0 x-{-7) , , I. , - 4 » ■ ■* ► i-7 3 (£) Figure 3-28 (Or we may think of Q as Xi and P as X2. Then the theorem gives * -5 0-5 equation obtained by the first method.) us -£ 2. — * ^ , which simplifies to x' = \(x + 7), the same 122 Distance and Coordinate System* Chapter 3 Example 3 Given three coordinate systems with points and coordi- nates as marked in Figure 3-27, express x in terms of Jfc. Then express x' in terms of k and, finally, express x' in terms of *. ABC ■ • • • — ► a s x o i ft ($') Figure 3-27 n 3 ,' (§-) Solution: (1) i-=4 * -^4 or * = ^ + 2 w 6-2 1-0 (2) £ - ~ n = * ~ ° or x 1 = -8k + 1 1 6 - 2 1 - - x' - 11 Jt- - 3- 11 1 - - x* - 11 _ x - - 2 (3) sfu =tr| or ^^-zt+is In working an example like this one yon may need to write more details than we have shown. A more complete version of (3), for exam- ple, might he as follows: 3- - 11 6-2 x* - - 11 -8 x-2 4 x' -11 = -%~ 2) x* = _2x +4+11 X* - -2ac + 15 Example 4 Given two coordinate systems with points and coordi- nates as marked in Figure 3-28, express x' in terms of x. 4 A B X Future 3-.2K 3 -I -a -10 IS) <£") Solutinu: rf - (-8) -10- (_8) 3^ + 8 -10 + 8 x' + S x- 3 -1-3 x- 3 -4 x-3 -4 3,6 Two Coordinate Systems on a Line 123 ^_U_!9 or *=»-"> 2 tSxantpte 5 Given two coordinate systems with points and coordi- nates as marked in Figure 3-29, find x. B x -• • != ► 10 19 (S> -23 -88 x (§') Figure 3-S9 Solution: *-{-23) 19-0 -38 - (-23) 10-0 x +- 23 _ 19 -38 + 23 B ' 10 *+23 _ t fi x + 23 = -28.5 * = -51.5 llxamph (i A train traveled at a uniform speed on a trip from Chicago to New Orleans. If it was 180 miles from Chicago at 7:00 p.m. and 320 miles from Chicago at 9:00 p.m., how far from Chicago was it at 10:20 p.m..' Solution! (See Figure 3-30.) Think of hours past noon as forming one coordinate system and miles from Chicago another one. Suppose the train is x miles from Chicago at 10j hours past noon, that is, at 10:20 PM. Then x- 180 10|-7 x- 180 3^ or 320 - 180 9-7 140 2 so x - 180 = 140(-~) = -i^, x = 180 + -292 = 413^. 6 6 3 1 1 g* (8) i 180 320 x Figure 3-30 "Hierefore the train is 413^ miles from Chicago at 10:20 p.m. 124 Distance and Coordinate Syittm* Chapter 3 Example 7 Given that 7. - 12, p are the respective coordinates of points A, B t P on a line/, that P C AB, and that AP a 3- AB, find p. Solution: (See Fjgu re 3-3 1 .) I n the coordinate system with A as origin and B as unit point the coordinate of P is 3, Then 4 • • • hi p -12 7 Figure 3-31 | 10 Ezample 5 Given that 7, - 12, p are the respective coordinates of points A, B, Pon a line I, that P £ opp AB, and that AP = 3 ■ AB t find p. Solution: (See Figure 3-32.) In the coordinate system with A as origin and B as unit point the coordinate of P is —3. Then p - ( - 12) —3-1 p + 12 7-(-uS = -ori- i§- = 4 - p + 12 = 7 «. p = 6 *- < — * — i £ M -12 7 p Figure 3-32 i o -4 EXERCISES 3.6 ■ In Exercises 1-5, a, h, c are the respective coordinates of points A, B, C on a line /. State which point, A, B, or C, is between the other two, 1. a = 0, b = 5, c = 100 2. a = a & = 5, c = - 100 3. a= -3,fc = -7,c = 7 4. a = 0.500. & = 0.O50, c = 0.005 5. a = J, h = i c = - j ■ In Exercises 0-15. a, fc, c- are the respective coordinates of points A , B, C on a line /. From the given information determine whether b < c or b > & 0, A-JB-C, o = 10, /> = 20 11. A-C-B, a = 8 t b = Q 7. A~#-£ a = 20, b = 10 12. B-C-A, a a 8, b = 8. B-A-C, a = 10, 6 = 20 13. lM^, a = - J, 6 = -0.66 9. B-A-C, a = 20, h = 10 14. B-C-A, a = §> b = 0.66 10. A-C-B, a = 0, b = 8 15, B-C-A, a m -|, h = 3.6 Two Coordinate Systems on a Line 125 In Exercises 16-20, a, b t c are the respective coordinates of points A, B, C on a line I. From the given information determine the number c. 16. A-B-Q a = 0,fc = l,AT5s*BC 17. A-C-B, a = 0, b = hW s*CB IS. B-A-C, a - 0, b = 1,AH sAC 19. A-B-C, a = -17, b = -B,XB==B€ 2ft A-B-C, a = -17, fc = 8, M m ~BC In Exercises 21-25, there is a sketch of a line with some points and coordi- nates marked. Find the coordinate x. ABC ,, R Q X 21. i » • • ► 24. 4 • •— « ► I 1 m 3 13 15 (S) 2*3 (§•) 1 x (S r ) X P Q A B P 22. 4] • • • ► 25. 4 • • • ► x 1 (£) 16 8 30 (g) 29 B -10 (g'J -1 8 i (g'J 23. 4 *■ 3 11 13 <$) i 1 (S) In Exercises 26-30, there is a sketch of a line with some points and coordi- nates marked. In each exercise write an equation relating x and x r and sim- plify it to the form x f = ax ■+■ h, (Note that a must be a number different from 0. For if a = 0, then x* = b and every point X would be matched with the same number in the system §'. Since §' is a coordinate system, we know tiiat different points must be matched with different numbers.) Check your answers by substituting the values of x at A and B in the equation to see if you get the corresponding values of x 1 at A and B. 26. 4 27. 2S, A I X 29. A X B 2 8 X x' fin 4 X x' 4 i A B 34% A B X X x' I & 5 -7 i» ■ 40 400 X x' (S) A X B ~3 9 X X* -10 IB 126 Distance and Coordinate Systems Chapter 3 ■ In Exercises 31-35, a subset S of a tine / is given in set-builder notation. Sketch the graph of the set and mark the ^-coordinates and ^-coordinates of three points of 5, Exercise 3 1 has been worked as a sample. 3L $= {X:x=6fc + 2 l *£2} Sftlution ion: S = {x:^-2 = jt,fc<2j * £ 1 X 14 8 2 32, S ss {X : x = 3k + 6, < k < 1} 33, S = {X : x = -3k + B, 0<k< 1} 34, S = {X : .*= -3* -8, * > 0} 35. S = (X : 3C — 3 - 7 - : k-0 I -0 2 14 + t , k is any real number} In Exercises 36-40, A and B are points on a line with coordinates 7 and 12, respectively. Enid the coordinate of the point F subject to the given condition, 36. F C AS and AP = 3 • AB 37. F £ AB and AP = j • A23 38. P € opp AB and AP = 3 ■ AB 30, P C AB and-gi- s 2 (two possibilities) FB 40, F e BA and AF = ±-PB (two possibilities) In Exercises 41-50, A r B,P are points on a line t with coordinates 0, 1, Jt, re- spectively, in a system § and with coordinates —3, 7, x, respectively, in a system §'. Figure 3-33 is an appropriate one for Exercise 41. Figure 3-33 AT a i (S) 7 <£') "M 41. II P £ AB and ^ = % find fc and *. AB 42. If F £ opp AB and-— a |. find fe and k 43. If It a -2. find AP AB P £ AB, or is F £ opp AB? 3.6 Two Coordinate Systems oci a Line 127 44, lfk = 5, find 41- 1* P € AB, or is P £ opp AG? AS 45. Ifx = 17, find A- and 44- 40. Ifx=27. find ^c and 4^- 47. lfx = 27, find^. 48. If* = 37, End^-. 49. lf* = 47, find^l-. 50. Hx = 0,find-^, 51. Given the situation of Example 6 on page 123, express die number x of miles from Chicago in terms of the number t of hours past noon of the day the trip began. 52. Given the situation of Example 6, find the "departure" time at Chicago. 53. On a car trip across the country Mr. X stopped at Ridgeville and "filled it up"; his odometer reading was 35378. Sometime later the gasoline gauge read | full and the odometer read 35513. Assuming that there is a constant "gasoline mileage," what does the odometer read when the gauge reads g? 54. Given the situation of Exercise 53, express the odometer reading m in terms of the amount x of gasoline in die tank, where x = 1 when the tank is full and x = when the tank is empty. 55. Mr. X reeendy completed a 180-day weight reduction program. If he weighed 200 lb. on the tenth day and 180 lb. on the 100th day, how much did he weigh on the 150th day? Assume thai he loses the same weight each day. 56. Given the situation of Exercise 5-5, express Mr, X's weight w [hi pounds) in terms of the number n of days, where n = 1 means the first day of the program, » = 2 means the second day of the program, and so on. 57. Coder certain standard conditions the freezing point of water is 0° Centigrade and 32° Fahrenheit; the boiling point of water is 100* Cen- tigrade and 212" Fahrenheit. What is the temperature in degrees Cen- tigrade when the Fahrenheit reading is 77"? 58. Given the situation of Exercise 57, let C denote the number of degrees Centigrade and F the number of degrees Fahrenheit. Obtain an equa- tion that relates Q and F by substituting appropriate numbers for ¥%, F*, P J? ("* f"* Cn Ca in the following equation: r ~ ' * =-?; tt • 128 Distance and Coordinate Systems Chapter 3 59. Starting with the equation obtained in Exercise -58, derive an equation in simplified form that expresses F in terms of C. 60. Starting with the equation obtained in Exorcise 58 } derive an equation in simplified form that expresses C in terms of F. ■ In Exercises 61-65, a line with points and coordinates as marked in Figure 3-34 is given. Using the Two Coordinates Systems Theorem, x' may be ex- pressed in terms of x, first with A as Xi and B as X&* and then with B as Xj and A as Xa- The results are * + 1 _ * - 2 «* *> - r_5 « i^f=T^ w 5-2 w -1-6 2-5 A B X 4 f * » 1 2 ■> x -1 6 x' Figure 3KM If A = A, then x = 2, r' = - I, and Equation (a) beeomes - y . , — - ~ n which reduces to = 0. o + 1 5 — 2 61. Simplify Equation (a) if X = & 62, Simplify Equation (b) if X' = A. 63. Simplify Equation (b) if X = B. 64, Tlie sum of the left members of Equations {a) and (b) h tf + 1 ar-^6 6+1 -1-6 which simplifies to * + i . g'-fl _ «* + i x'-e _* + i - (x- - 6) 7 . 7 + -7 _ 7 f~" 7 = 7 = L Add the right members of Equations (a) and (b) and simplify. 65. If you multiply the left member of Equation (a) by — 1 and add 1 to the product, the result is x + * • (-1) + L This simplifes to y +J . ^j , T _ -* - 1 7 7 1 + - " 7 + 7 _ -x + 6 x' - 6 _ x / - 6 7 : ~7 : -1-6' which is the left member of Equation (b). Multiply the right member of Equation (a) by — 1 and add 1 to the product. Show that the result sim- plifies to the right member of Equation (b). 3.7 Point* of Division 129 3.7 POINTS OF DIVISION In I his section we use the Two Coordinate Svs terns Theorem b i help us find the coordinates of the points on a given segment which divide it into a given number of congruent parts or divide it in some other specified way. Definition 3.S The midpoint of a segment AB is the point P on AB such that AP = PB = \AB. The midpoint of a segment is said to bisect the segment or to divide it into two congruent parts. Definition 3.7 The trivection points of a segment AS are the two points F and Q on AB such that AP = PQ = QB = The trisection points of a segment are said to divide the seg- ment into three congruent parts. Similarly, points C. D, and E on "KB such that AC=CD = DE=EB = ±AB are said to divide AB i nto four congruent parts. This idea may be extended to any number of congruent parts. A segment AT? is a set of points. You might think of a "path" from A to B if you were to draw a picture of the segment. Although you may think of AB as a path, be careful to remember that the segment from A to B is the same as the segment from B to A, Indeed, AB = BA, Sometimes, however, we want to consider B to A as different from A to B. It might be helpful to think of trips and to consider the trip from A to B as different from the trip from B to A. The point which is one-third of the way from A to B is different from the point which is one-third of the way from B to A. This leads us to the idea of a directed segment. We think of a directed segment as a segment with one end- point designated as the starting point. A directed segment is known as soon as the segment is known and the starting point is known. Our formal definition follows. 130 Dfst«nc« and Coordfnate Systemi Chapter 3 Definition 3.8 The directed segment from A to B, denoted by AR is the set {Al, A}. It is important to note the difference between the symbol for a rav the symbol for a directed segment The ray symbol, as in AB a has a complete arrowhead, whereas the directed segment symbol, as in AB, has a half arrowhead. Directed segments are related to vectors, and half arrows are frequently used in vector notation, Vectors are verv useful in many branches of higher mathematics. Note that whereas AB = M t it is not true that AB - BA. Note that whereas = (AB,A) = {BA i A} t = {/U?, B} = {HA.BJ. Definition 3.9 Let a directed segment AB and two points P and Q on AB be given. If P £ AE, Q g AB y and AP AO PB ~ OB * ^ shown in ^ i § ure 3 "35, tnen ** and (J are said to divide AB in the same ratio, P dividing it internally and called an internal point of division, Q dividing it externally and called an external point of division. The ratio ^- is the PB ratio of division. A Figure 3-35 f B Example 1 Given two points A and Bona line I with coordinates 4 and 20, respectively., as indicated in Figure 3-36, find the coordinate of the point P on AB if P divides AB into two congruent parts. 20 <g> -► I Figure 3-36 3.7 Points of Division 131 Solution: Let x be the coordinate of the desired point P. Then 4 < * < 20, x - 4 s 20 - x, 2x = 24, and s = 12, Alternate Solution: Set up two coordinate systems as indicated in Figure 3-37. P B kf -# •- — H 4 * 20 (S> Figure $3f The i . -4 _ fe- -4 " 1-0 and ar = 4+ 16A: for every P on i*. Then F divides A5 into two con- gruent parts if k = ^ and x = 4 + 16 • ^ = 4 + 8 = 12. Example 2 Given two points A and B o n a line I with coordinates — 3 and 21, respectively, find the coordinates of the four points which di- vide it into five congruent parts. Solution: Let S and $' be two coordinate systems on J with coordi- nates of several points as marked in Figure 3-38. A P B 3 ! 3! 55 — M * i ($'» Figure 3-3S Then x- (-3) _ x+ 3 __ fc-0 21 _ (_3) :: 24 1 -0 = Jt and x — 24* — 3 for every point P on /, The required points are the points with ^-coordinates \-> |» |, y or. in decimal form, 0.2, 0.4, 0.6 t 0.8. Now we compute the a:- coordinates of the division points using the equation x = 24k — 3. They arc 24(0.2) - 3 = 4.8 - 3 = L8. 24(0-6) - 3 = 144 - 3 = 11.4. 24(0.4) - 3 = 9.6 - 3 = 6.6. 24(0.8) - 3 = 19.2 - 3 = 16.2. 132 Distant* ami Coordtnite Systems Chapter 3 Example 3 Given two points A and B on a line I. with coordinates 3 and -21, respectively, find the coordinates of the points P and Q on AB which divide AB internally and externally, respectively, in the ratio Z. Solution: Let $ and g' be two coordinate systems on I with coordi- nates of several points as marked in Figure 3-39. Q A p b T7. * * * ■ — H {&) ** 3 * -21 Figure 349 ($') ** 0*1 Then, as in Example 2, ^J- = k and x s -24k + 3 — 24 AQ 7 for every point F on J. Since -^ = - f A{> is less than QB and ^ is a point on opp AB. The coordinates x and a:' of F and Q are computed as follows: AF=k AQ = - ^ PB=1 -k p£ = 1 - if AF 7 k A() 7 -f pB 8 "' 1-tf FB 8 1 - k r- Ik s Sfc 7 - 7Jf = -8Jf 15* = 7 fifs -7 la **= -24(^7) + 3 = 171 ,= _24.X + 3 = -8.2 Check (using r-coordinatesK AF 3-(- PB -8.2 - -8.2) (-21) 1L2 112 7 " 12.8 " 1.28 " 8 AQ 171 Qff 171 - { 3 -21) K>S 42 7 192 ~ 48 ~ = 8 Example 4 Given two points A and B on a line I with coordinates 3 and -21, respectively, find the coordinates of the points R and Ton AB which divide BA internally and externally, respectively, in the ratio -J. (Compare with Example 3.) 3.7 Writs of Division 133 Solution: Let $ and $' be the two coordinate systems on I with co- ordinates of several points as marked in Figure 3-40* Then, as in the preceding examples, * - +" ^ = k and x — 24fc — 21 for ever}' point HT 7 > R on /. Since ~f = ^, BT is less than 7"A and T is a point on opp BA. /A 8 < i R B T 4 r i X it -21 Figure 3-40 Then BR = k BT = - V RA = 1 - k TA = 1 - K BR __ 7 _ fc BI 7 -jtf flA "" 8 I - k TA" S 1 - Jf k-4- 15 (Why?) £ = -7 (Why?) x = -9.8 (Why?) s? = -189 (Why?) Check (using x-coordinates). Bfl _ -9,8 + 21 _ 11.2 _ 112 _ 7 RA 3 + 9.8 12.8 128 ~ 8 BT _ -21 H- 189 ^ 168 = 1 TA 3+189 EXERCISES 3.7 In Exercises 1-5, A and B are points on a line I with given coordinates a and &, In each exercise, find the coordinate of the midpoint of ~KB. L o - 5 S b = 27 4, a = S, b = -27 2, a = -5, h = 27 $. a = 0*b= -4.8 3. a = -5, h = -27 In Exercises 6- JO, A and B arc points on u line / with given coordinates a and b. In each exercise, find the coordinates of the points which divide AB into the given number, n, of congruent parts. 6. a = -3, b - - 7, n = 2 9. a = 8, b = -8, ft = 8 7. o = 3, b = 0, « = 5 10. a = 0, b -= 10, n = 4 8. a = -], h = 79, n = 3 134 Distance and Coordinate Systems Chapter 3 ■ In Exercises 11-15, A and B are points with given coordinates a and b. In each exercise, find the coordinates of the points P and Q which divide A~B internally and externally in the given ratio r. 1L a = 10, b = 20, r = I 14. a = f, b = $, r = ^ 12. a = 20, fr = 10, r = ? 15. a = |, fc = J. r = ^ 13. a = 26, fc = 0, f = f 16. Prove the following theorem. THEOREM If tine coordinates of A and B are « and h t then the coordi- nate of the midpoint of AH is a + h . 17. Prove the following theorem. THEOREM If the coordinates of A and B are n and h, then the coor- dinates of the bisection points of AB are ^ + b and - + ^ . . 3 3 18. A and B are points on a line / with coordinates and 1, respectively. Find the coordinate of the point P which divides AB externally in the ratio -$&, 19. A and B are points on a line / with coordinates and 1, respectively. Find the coordinate of the point Q which divides AB external] v in the ratio -tf$L, 20. A and B are distinct points on a line /. Is there a point P on AB*, but not on AB, snch that P is the same distance from A as it is from B? That is, is there a point P on AB which divides AB externally in the ratio y? ■ In Exercises 2 1 -26, A and B are points on a line ( with given coordinates a and b, respectively, P is the point which divides AB internally in the given ratio r. la each exercise, select the statement from the right-hand column that is true. 21. a = 0, b = 1, r = ] (A) P is between the midpoint of A~E and # 22. a = 0, b = 10, r ■ { (R) AP = 1 23. a = 10, b = 209, r = -$$ (C) P is a bisection point of AB 24. a e 10, b = 211, r e jgj (D) BP = 1 25. a = - 1.3, fc = 0,ra-^ (E) P is the midpoint of OT 26. « = 0, h = 13, r = J£ (F) P is hetween the midpoint of ^ and A Review Exercises 135 ■ In Exercises 27-30, A and B are points on a line I with coordinates a and h, respectively. P is the point which divides A externally in the given ratio r. In each exercise, select the statement from the right-hand column that is true. 27. a = Q.b = 1, r = ffifa (A) F-A-8 and BP = 1000 2S. a = (Kb=z hr= jggi (B) A-R-P and HP = 0,001 S8L Os^sUa U ^ L (C) A-B-P and BP = 1000 30. a = 0, h = 1, r = -^ (D) P-A-tf and AP = 0.001 CHAPTER SUMMARY In this chapter we have developed the concept of DISTANCE between two points and the concept of a COORDINATE SYSTEM on a line. We mtroduced five postulates: DISTANCE EXISTENCE POSTULATE, DIS- TANCE BETWEENNESS POSTULATE, TRIANGLE INEQUALITY POSTULATE, DISTANCE RATIO POSTULATE, and RULER POSTU- LATE, This chapter contains many definitions and theorems. A key defini- tion is the definition of a coordinate system. The climax of the chapter is the TWO COORDINATE SYSTEMS THEOREM. In Section 3.7 we applied the tools of this chapter to find points of division which divide a segment internally and externally in a given ratio. REVIEW EXERCISES In Exercises 1-5, name the property that justifies the given statement. 13+4=4+3 2* (AB + CD) + KF = AH + (CD + EF) 3. If a — b> then 7? = a. 4. (.50 + 5) • 3 = 50 ■• 3 + 5 • 3 5. (5 + 3)-7 = (3 + 5)-7 In Exercises 6-11, name the postulate which justifies the given statement. 0, If P, Q, R are distinct collmear points with P-Q-R, then PQ + QR = PR. 7. If X and Y are distinct points and AH is any segment, then the distance between X and Y in the distance function based on A2J is a positive number. 8. If M, W> K arc three noncolliuear points, then RK + KW > RW. 136 Distance and Coordinate Systems Chapter 3 9. If A, B, C, D, E, F. G t H arc eight distinct points* then AB (in ~EF units) _ AB (in C?r7 nnfts)_ CD (in EF units) CD (in C77 units) ' 10. If A, ii, C, D t E, F, G, Ware eight distinct points, then AB (in AT units) CD (in Eb units) AB (in C77 units) " CD (in CT units)* 11. If A, B, C t D are four distinct points, then there is a unique coordinate system on A B relative to CD such that the origin is B and the unit point is A. ■ In Exercises 12-20, A, B, and X are points on a line h a, h, % are their re- spective coordinates in one system j a\ h\ xf are their respective coordinates in another system. In each exercise, (a) draw and label an appropriate figure, (b) express x' in terms of x and simplify, and (c) express x in terms of x 1 and simplify. 12. a a 0, b = 1, tf = 5, V = 8 13. a = 0, b = 1, d = -8, V = 5 14. c= 1, b = 0. a' = 5, tf = 8 15. a = 17, b = 16. a' = 5, &' = 8 16. o = 7, b = -3, ^ = 0, 6' = 1 17. a = 0, fc = 1, a' = 0, &' = 2 18. a = 0, b = 1, </ = a V = -1 19. a = 0, fo = 1, a' s 1, ¥ = 20. a = 0, /j = 100, «' = 1, fo' = ■ In Exercises 21-30 t A, B, C f D. E, are points on a line I with coordinates 0, 1, 2. 3, 4, respectively, in a coordinate system on I In each exercise sim- plify the given expression. Each expression names a number. 21. AB (in AB units) = [T] 22. AB (in AT unite) = \f] 23. AC (in ATJ units) = [TJ 24. AD (in Al units) = (TJ 25. AD (in AC unite) s= [?] 26. AD (in A~E unite) s [?] 27. AE (in AB unite) = {T\ 28. AB (in A~E unite) = 29. Ag(inXEui U ts)_ AD (in AT? unite) 30 AB (in IC unite) _ — " AD (in AT unite) U Review ExerciiB* 137 In Exercises 31-40, A$ is a directed segment, P divides AB internally in the positive ratio - , Q divides A~fi externally in the ratio - , and a and h are the coordinates of A and B, respectively, in a coordinate system cmAB. In each exercise, draw un appropriate figure und find p and q, the coordinates of P and Q t respectively. 31. o = 0,ft = l,f=8,ia 1 36. a = 10. fe = 5.-^ = 4 32.a = 0=l,r:=I p *=2 37. a = -G t fc = 9, - = -i- i' o 33. a = 0, h = 1, r = 2, a- = 3 38. a = -6, b = 9, - = 3 34. a = 0. h = 1, r = 10, s = 5 39. a . = 1, b = 2, r = 5, * = 6 35. a = 10, h = 5, J- a 4" 40. a = 1.3, & ss 7J*- a A s 4 * oo In Exercises 4 1-47, X is a point on a line / and x and k are its coordinates in two coordinate systems on I. The given equation tells how x and k are related for every X on L In each exercise, copy and complete the given statement. 41. x = 3fe + 1 If k - 0, then x = 42. x = 3*+i If fc= 1.thenx = |7] 43. x = 3fc -+■ I < Jt < 1 if and only if fj] (a condition on x) 44.x=-3Jt+l If k = -3, then a = [T| 45. x m -3Jt +1 If Jt = -5. then x = [?] 46. a: = — 3fc -f 1 k < — 3 if and only if JT] (a condition on x) 47. i = — 3fc + 1 fc > — 3 if and only if \T\ (a condition on i) Fred Ward /Black Star Angles, Ray-Coordinates, and Polygons 4.1 INTRODUCTION In Chapter 3 we developed definitions and postulates for the con- cept of distance and for coordinate system* on a line. Bctweenness for points is related to betweenness for red numbers through the idea of a coordinate system on a line. The length of a segment is related to the coordinates of its endpoints. In this chapter we develop postulates and definitions for angular measure or, as we usually call it, angle measure, and for ray-coordinate systems in a plane. Betweenness for rays is related to bctweenness of the coordinates matched with the rays. The measure of an angle is re- lated to the coordinates of the rays that form the angle. Ray-coordi- nates are useful in developing the properties of angles. Chapter 4 concludes with a discussion of polygons and dihedral angles. The idea of a polygon is a natural extension of the idea of a tri- angle, and the idea of a dihedral angle grows naturally from the idea of an angle. 140 Angles, Ray-Coordinates, and Polygons Chapter 4 4.2 ANGLE MEASURE AND CONGRUENCE sen a pie is cut into four quarters of equal size as indicated in Rgure 4-1, the rim is also cut into quarters. This is true regardless of the size of the pie. If a pie is cut in the usual manner, then with each piece that is less than half a pie there is an associated angle, sometimes called the associated central angle. This angle is the union of the two mys which have the point at the center of the pie as their common end- point and which contain the segments that are the cuts forming the piece of pie. For a quarter pie w r e might think of the size of this angle Figure 4-1 This corresponds to thinking of a revolution as the unit of meas- ure. If we adopted this unit, then the measures of the angles in our ge- ometry would be real numbers between and j. We prefer to think of one revolution as equivalent to 360 degrees. Then the measures of angles in our formal geometry will be numbers between and 180 as suggested in Figure 4-2. Figure 4-2 It is said that the Babylonians originated the system of measure- ment that is based on what we now call the degree as the unit. To them the stars (except the sun) appeared to be fixed on a celestial sphere that rotated about an axis once each day. The sun appeared to com- 4.2 Angle Measure and Congruence 141 plete a circular path among the stars once each year (four successive seasons}. They apparently knew that the length of the year was approx- imately 365 (lays, but, perhaps for convenience* took 360 days as their "calendar" year. Considering that the sun traveled over a circular path once each 360 days it was natural to divide that path into 360 equal parts and consider each part as corresponding to one day and 90 parts as corresponding to one season. In the early days of the Christian era, the Greek mathematicians of the School of Alexandria divided the circle into 360 equal parts and called each part a moira* This Greek word was translated into the Latin word de-gradus, meaning *'a grade or step from/' From this we get our word degree, meaning the first step down from a complete revolution, or -g£$ of a revolution, Of course* we could use other units of angular measure such as rev- olutions or right angles. Some units that you may not have heard of are mils, grads, and radians. There is no particular reason for using degrees for angle measures other than the fact that this is commonly done and lias been done for a long time. To make our development simpler, we base our formal geometry of angle measure on just one unit of measure, the degree. POSTULATE 20 (Angle Measure Existence Postulate) There exists a correspondence which associates with every angle in space a unique real number between and 180. Definition 4.1 The number which corresponds to an angle as in the Angle Measure Existence Postulate is called the measure of the angle. Notation. The measure of A ABC is denoted by m L ABC. Note that if the number of degree units in the measure of I. ABC is 40, then ml ABC = 40 and not mLABC = 40*. If an angle is marked 40° in a figure, it means that the measure of the angle is 40, In Chapter 3 we agreed to call two segments congruent if they have the same length. We make a similar agreement for angles. Definition 4.2 Two angles (whether distinct or not) are congruent angles, and each is said to be congruent to the other if they have the same measure. Notation. Z ABC s? ADEF denotes that I ABC and IDEF are congruent. 142 Angles, Ray-Coordinates, and Polygons Chapter 4 Although congruence and equality as applied to angles may seem alike, they arc in reality different ideas. It is true lhat if LA = LB, then LA ^ LB r but the converse is not true. Many pairs of congruent angles are not pairs of equal angles. Remember that an angle is a set of points and that two sets of points are not equal unless they consist of the same point*;. For the angles suggested in Figure 4-3 we have LABC = L CBA = LARK = L GBC = L GBK. Figure 4-3 Also, LABC^ LDEF. Rut it is quite possible that mLARC = mLDEE If this is true, then LABC m L DEE The most common device for measuring, angles in informal geom- etry is a semicircular protractor with degree marks from to 180 evenly spaced on the semicircular edge. To measure an angle such as in Figure 4-4a we can either place the protractor as indicated in (b) and read the measure 20 directly or vvc can place it as indicated in (c) and ob- tain the same measure by subtracting 65 from 85. Pbpus I- 4 Another type of protractor is a circular one. (See Figure 4-5.) This 380-degree protractor has advantages in drawing certain figures. In using a 360-degree protractor, as in using a semicircular one, it is pos- sible to obtain the measure of an angle from readings on its scale. This and other properties of the protractor suggest the concept of a ray- coordinatc system which we define later. 4,2 Angle Measure and Congruence 143 Figure 4-5 EXERCISES 4.2 1. Copy and complete the following definition of congruence of angles. If Z A = L B, then [fj and if »?£ A = mZB, then |TJ. 2. (a) Use your protractor to construct three angles whose degree meas- ures are ©0, 90, and 135, respectively, (b) Why is it not possible, using our definition of angle measure, to con- struct an angle whose degree measure is 240? 3. Use your protractor to find the degree measure of the angles shown. 144 Angfas, Ray- Coordinates., and Polygon* Chapter 4 4. In Exercise 3, did you find two angj.es thai arc congruent? If so, name them and tell why they are congruent. 5. If Sandy Moser measures an angle and finds its. degree measure Lo be 60 and Bob Blake measures the same angle with the same protractor and finds its degree measure to be 70, what can you conclude? What postulate are you using as a basis for your conclusion? Exercises 6-10 refer to the angles in Figure 4-6, 4 iur E U 1KT rss R B Figure 4*0 6. Using the notation of the figure, write two different names for equal angles, that is, for the same angle. 7. Name two angles that arc congruent but not equal. 8. Write two names of angles such that the angles named are equal (and therefore congruent). 9. Name two angles that are not congruent. 10, Can you name two angles that are equal but not congruent? 11. Using a protractor, measure the angles of the two triangles in the fol- lowing figure. List those pairs of angles that appear to he congruent 4.2 Angle Measure and Congruence 245 12. In the figure, the readings for certain rays with endpoint V are shown in a circular protractor. Name four pairs of congruent angles, that is, name four pairs of angles such that the angles in each pair are congruent to each other. 13. Without using a protractor draw six angles whose degree measures you would estimate to be 30, 45, 60, 90, 120* and 150, respectively. Alter you have drawn each angle, measure it with your protractor and see how good your estimate was. 14. Draw a triangle ABC so that m Z A = 43 t mlB = 57, and m Z C = 80. 15. Can you draw a triangle DEF so that mlD = 54, m£E = 67, and mZF = 70? 16. Draw a triangle DEF so that mi. D = 54 and m/.E = 67. Use your protractor to find m L F, In Exercises 17 and 18, complete the proof of the following theorem. THEOREM Congruence for angles is reflexive, symmetric, and transitive. Proof; Let Z A be any angle; then in Z A is a number and mAA = m Z A by the reflexive property of equality. But if mLA = m/_A t then LA^ L A by the definition of congruence for angles. Therefore congru- ence for angles is reflexive. 17. Prove that congruence for angles is symmetric 18. Prove thai congruence for angles is transitive. 1». (a) If LAm LB and LC^ IB, what can you conclude? (b) What properties justify your conclusion in (a)? 146 Angles, Ray-Coordinates, and Polygons Chapter 4 4.3 BETWEENNESS TOR RAYS Recall that if A, B, C are three distinct points on a line, then B is between A and C if and only if AB+ BC = AC Also, B is between A and C if and only if B is between C and A. Be- tweenness for points is related to betweenness for "numbers througji the definition of a coordinate system on a line and the Ruler Postdate. For three distinct coplanar rays VA t VB, VC (with a common end- point) we want to develop a concept of betweenness based on our in- tuitive notions of symmetry, our experiences with protractors, and our desire for additivity of angle measures in certain situations. Specifically, if V£J is between V'A and VC, then we want VB to be between VC and VA, Also, we want VB to be between VA and VC if and only if m Z AVB -r m£BVC= m I AVC These ideas suggest the following definition and postulate. Refer to Figure 4-7 us you read them. ix , jfr-f , J&l A and B are on tha / B and C are on the V A and C are on same aide of VC / aanw aide of VA opposite sadea of VB m V r i i» betwaan V~X and V# Figure 4-7 — ♦ — ► — > — > Definition 4.3 If VA, VB, VC arc rays, then VB is between VA and VC if and only if 1. A and B are in the same h airplane with edge VC. 2. B and € are in the same half plane with edge VA. 3. A and C are in opposite half planes with edge VB. POSTLXATE 21 (Angle Measure Addition Postulate) If VA, VB, VC are distinct coplanar rays, then VB is between VA and VC if and only if mLAVC = mZAVB + tnZBVC 4.3 Betweenness for Rays We consider two important matters relating to Definition 4.3. (A) Suppose that VJ? is between VA unci VC. Is VB between VC and V?? (B) Suppose that V$ is between vX and V&, and that A', B', C are any points, except V,.on VA, VB, VC, respectively, as in Figure 4-8. Is vS' between vA' and VC?? In view of Definition 4,3, the questions raised in (A) and (B) amount to the following, expressed in terms of Figure 4-8. Suppose (1) A and B arc in the same 147 Figure 4-S half plane with edg<& VC, (2) B and C are in the same half plane with edge H, (3) A and C are in opposite half planes with edge VB. Does it follow that (4} C and B are in the same halfplane with edge VA, (5) B and A are in the same halfplane with edge VC, (0) C and A are in opposite halfplancs with edge VB? Does it follow that (7) A' and B' are in the same halfplane with edge VC, (8) & and C are in the same halfplane with edge VA\ (9) A' and C are in opposite halfplanes with edge VB'? Since (1) implies (5) and (7), (2) implies (4) and (8), and (3) implies (6) and (9), it follows that the answer to questions (A) and (B) is Yes. Thus betweenness for rays is symmetric just as betweenness for points is symmetric. Indeed. VB is between VA and VC if and only if VB is between VC and VA, and Q is between P and R if and only if Q is between H and P. Also, betweenness for rays depends on rays, not on the particular choice of points used in designating the rays. Postulate 21 is consistent with these properties of betweenness. Thus, referring to Figure 4-8. nUAVC = mlAVB -j- mlBVC mlCVA = mACVB + mlBVA and m£AVC = m£AVB + mLBVC mLA'VC = mlA'VB' + m£B'VC if and only if (Why?) if and only if (Why?) 148 Angles, Ray-Coordinates, and Polygon* Chapter 4 Example Consider the six coplanar and concurrent rays formed by the three intersecting lines in Figure 4-9. Then study the following two instances of betweenness in this figure and the four instances of "not betweenness," Betweenness 1. VA is between VFand VB since (a) B and F are on opposite sides of VA, (b) A and F are on the same side of VB. and (c) B and A 4 — ► are on the same side of VF, From Postulate 21 it follows that ml FVA + ml AVB = mlFVB. 2. VA is between VB and VF. Indeed, the three statements to check are the same three statements that we just checked to verify that VA is between VFand VB, From Postulate 21 it follows that mlBVA + mlAVF=mlBVF, which should not be surprising in view of the equation in L FVA + m L AVB = m I FVB and the commutative prop- erty of addition, Not Betweenness 1. VA is not between VF and VC, Why? Let us check. Are F and C on opposite sides of VA? Yes, Are A and F on the same side of VC? No. Of course, one '"So" in checking the three require- ments is enough to establish "not betweenness." An alternate method of checking betweenness in this instance in- volves using Postulate 21. According to this postulate, VA is between VF and VC if and only if mlFVC = mlFVA + mlAVC 4.3 Betweenness for Ray* 149 But this equation is a false statement. Why? It is false because the right side of the equation is a number, whereas the left: side is not. Indeed, there is no such angle as Z FVC and hence there is no number such as ntlFVC, (See Section 4.2.) 2. VA is not between V? and VB. Why not? 3. VX is not between V7?and vd, Why not? 4. VA is not between VA and VB. Why not? EXERCISES 4.3 L In the figure below at the left, explain why ST is not between BA and EC A 2. In the figure above at the right, explain why VCis not between VA and ~VB. 3, In the figure for Exercise 2, is VB between VA and VC ? Is VA between V$and VC?? 4. In the figure at the right, if A-B-C. then is BD between BA and BC? Why? D 4 — * A G Exercises 5-9 refer to Figure 4-10, Assume that no two of the angles of the figure are congruent. In each exercise, name the missing angle. 5. m Z AEB + ml EEC = m Z \j] a mlAEC + ml CRD = mZ[T| 7. ml ABC - ml ABE = ml^} 8. ml BED — ml\?} = mlBEC 9. mZ[0 - mlECD = mlECB Flgur* 440 ISO Angles. Ray-Coordinates, and Pofyf on* Chapter 4 ■ Exercises 10-13 refer to Figure 4- 11 . Name the missing angle or number. Figure 411 A 10. mlADE + nuLEDC = mZ(7] 11. mlDAB - nUDAC = mZ0 12, DE + EB = JTJ 13, AC -AE = \7] 14 Given mlAVB = 35 mlAVC= 115 find mZBVCif (a) VB is between vA and vS, and (b) VA is between V7? and V?. 15. If t in a plane, mlAVB = 70 and mlBVC = 44, find mLAVC. Is there just one possible value for m £ AVC? Illustrate with a figure. 16. In the figures below. tnLABC- m/.EFG = 70 and mlDBG = ml HFC = 25. Prove thai LABD^ lEFll 17, In the figures below, mlDEF s mLHKM = 120 and m/_GEF mlHKN = 35, Prove that IDEG s ANKM, 18. In the figures below, mlPQS = mlDBC = m and mlSQR mlABD = 50. Prove that IPQR ~ ZABC. 4.3 Betwa«nn*u for Hays 151 Tn the figure below, mZl = mZ3 and mZ2 = mZi Prove that <LDEF = £DGF, ). challenge pnoni jcm. Let VB be between VA and VC and let R<p be between /i? and /IS as shown in die figure below. It appears that all three of the following statements might be true: (a) Z CVB s Z SK^>, (b) Z BVA at Z ^>HP, and :c) Z C VA a / SRP. Prove that if any two of these three statements are true, then the third one is also true. Hint: There are three things to prove. (A) If(a)aml(b)»then(c)- (B) If (a) and (c), then (b), (Q If (b> and (c), then (a). Proof of {A): There is a number r such that m Z CVB = m / SRQ = r, Why? There is a number $ such that m / BVA = m Z £ilP = s. Why? Then miCVA = r+i= m L SRP> Why? Then ZCVAssZSRP. Why? Now write proofs of (B) and (C). 152 Angles, Ray-Coordinates, and Polygon* Chapter 4 4.4 RAY-COORDINATES AND THE PROTRACTOR POSTULATE In the same way that the Ruler Postulate provides us with a mathe- matical, or abstract, ruler for assigning coordinates to points, \vc want a mathematical protractor for assigning ray-coordinates to rays. Wc first define what is meant by a ray-coordinate system. Then we adopt the Protractor Postulate which amounts to an agreement that ray- coordinate systems do exist and that they are unique if we pin them down in certain ways. The definition is based on experiences with protractors. Definition 4A (See Figure 4-12.) Let Vbe a point in a plane «♦ A ray- coordinate system in a relative to V is a one-to-one correspondence between the set of all rays in a with endpoint V and the set of all real numbers x such that < x < 360 with the following property: If numbers r and s correspond to rays VJR and VS in a, respectively, and if r > s, then mZRVS = r — * m£RVS = 360- (r-s) VR and VS arc opposite rays if r-s<180 if r-s>!80 if r - s = 180. Frgure4J2 Definition 4.5 The number that corresponds to a ray in a given ray-coordinate system is called the ray-coordinate of that ray. The ray whose ray-coordinate is zero is called the zero- ray of that system. Notation. We use cd VX as an abbreviation for ray-coordinate of Va. 4.4 Ray Coordinates 153 Example 1 Figure 4-13 suggests a ray-coordinate system where the ray-coordinates of several rays are given. For this example we have: mAAVR = 70 - = 70 mlAVH' = 360 - (250 - 0) = 110 mZPVR =90-70 = 20 mZ iW = 250 - 90 = 160 mlFVA = 360 - (270 - 0) = 90 VR = opp vR' since 250 - 70 s 180 VA and VA' are opposite rays since 180 — = 180 Figure 4-13 POSTULATE 22 (Protractor Postulate) If a is any plane and VA and VR are noncolHnear rays in a, then (1) there is a unique ray-coordinate system § in « relative to Vin which cd vX = and <?<J VB = roZ A VB and (2) if X is any point on the B-side of VA, then cd VX (in g) = roZAVX Example 2 Figure 4-14 shows a line VA and two rays vS and V? with jB and X on the same side of VA.. V A Figure 4-14 If ml_ AVE = 50. there is a unique ray-coordinate system in which cd vZ = and cd VB = 50, If mlAVX = 105 and X and B are on the same side of VA, then cd VX = 105. 154 Angles, Ray-Coordinates, and Polygon* Chapter 4 We now proceed to prove several theorems using our postulates aibout angles. THEOREM 4 J If Z A VB is any angle in a plane a and if £ is the ray-coordinate system in a relative to V in which cd VA = and cd VB = m Z -A VB, then the ray-coordinate of VX is (1) if VX = VZ* (2) 180 if f£ s= app vX. (3) between and 180 if X is on the B-side of VA. (4) between 180 and 360 if X is on the not-B-side of VA. Proof: Figure 4-15 consists of four parts which correspond with those of the theorem. V A CD Figure 4-15 1. In a ray-coordinate system there is only one number matched with each ray. Since VX = VA and since is matched with VA. then is the number matched with VA. 2. According to the definition of a ray-coordinate system two rays are opposite rays if and only if their ray -coordinates differ by 180. Since cd VA = 0, since VA = opp VA, and since every ray-coordinate is either or a positive number less than 360, it follows that cd V? = 180, 4— * > 3. Since X is on the B-side of VA, then cd VX = mLAVX. Since m Z AVX is a number between and 180, it follows that cd VX Is a number between and 180. 4. Let X' be a point such that VX and VX' are opposite rays. Since X is on the not-B-side of VA, it follows that X' is on the B-side of VA. Then cd VX' is a number between and 180 by part 3, and since cd VX is a number between and 360 which differs from cd VX' bv 180. it follows that cd VX is between 180 and 360. 4.4 Ray-Coordinates 155 The argument in part 4 expressed in symbols consists of the fol- lowing steps: 0<cd VX<360 < cd VX' < 360 < cd V?' < 180 cdVX- cd VX' = 180 cd V$ = cd V?' 4- 180 180 < cd V? < 360 In working with a protractor we know that if we are given any an- gle, say Z DEF, and a ray VA on the edge of a halfpkne 3C, then we can draw a ray Vfi with B in 3C so that IAVB = Z DEF. Our postu- lates permit us to do this in. the abstract as the following theorem suggests. IHEOREM 4,2 (Angle Construction Theorem) If Z DEF h any angle, if VA is any ray, if 5C is any halfplane with edge VA, then there is one and only one halfline VR in 3C such that /.AVBst I DEF. Proof: (See Figure 4-16.) Let us suppose that I DEF, VA, and 3C are given as in the statement of the theorem. Let a be the plane that eon- tains ;1C and let F be any point in 3(1, Let § he the unique ray-coordi- nate system in o relative to V in which cd VA = and cd VP = mlAVF. E p Figure 4-16 Let m Z DEF = b. Then < b < 180 and there is exactly one ray VB with B in 3C such that cd VB = b. Why? Then ml AVB = b = ml DEF and VB is the unique halfline in 3C such that LAVB^ Z DEF. The following theorem relates betweenness for rays and between- ness for coordinates'. 156 Angles, Ray-Coordinates, and Polygons Chapter 4 THEOREM 4,3 If a ray-coordinate system in which cd VA = 0, cdVB = b,cdVC = c with c < 180 is given, then VB is between VA and VC if and only if b is between and c. Proof: Figure 4- 1 7 stipes ts the "given' ' or hypothesis of our theorem , We prove two things, L If b is between and c, then VB is between vZ and v£ 2. If VJ?is between vA and v3, then h is between and c. Proof of 1: In-;, ,iv -t-17 Statement L e < 180 2. c > 3, < b< c < 4. mZAVB= 6-0 5, m/_BVC=c-b a mZAVC = c - 7, (b - 0) + (c - b) = c - 8, ml AW + mZBVC = mZAVC 9, VB is between VA and VC, Iteof-afSt Statement 1. VB is between VA and vS. 2. VA, VB, VC are distinct rays. 3. B is on the C-sidc of VA 4. 0, b* c are distinct num- bers. 5. < c < 180 1. Hypothesis 2* All ray-coordinates are non- negative numbers, 3. Hypothesis and steps 1, 2 4. Definition of ray-coordinate system 5. Definition of ray-coordinate system 6. Definition of ray-coordinate system 7. Properties of real numbers 8. Substitution (steps 4, 5, 6, and 7) 9. Angle Measure Addition Postulate L Hypothesis 2, Definition of betweenness for rays 3* Definition of betweenness for rays 4. Definition of ray-coordinate system and step 2 5. Hypothesis 4.4 Ray-CoordtoitM 157 6. < h < 180 7. m£AVB = b-0 mLAVC = c - 8, mZAVB + mLBVC mlAVC 9, 6 + mZBVC = c 10, mZBVC>0 11, fc < c 12, 0<fc<c 6. Steps 3 f 4, 5 and definition of a ray-coordinate system 7. Definition of a ray-coordi- nate system 8. Angle Measure Addition Postulate 9. Substitution (steps 7 and 8) 10. Angle Measure Existence Postulate 11. Steps 9 and 10 12. Steps 6 and 11 THEOREM 4.4 (Angle Measure Addition Theorem) If distinct rays VB and VC are between rays VA and VD and if Z AVB s ZCVD, then LAVC sa LBVD. ' Proof: Figure 4- J 8 suggests two possibilities. We prove both cases at the same time. v a v a F*w*4-lS (.) (b) Suppose that VB and VC are distinct rays as in Figure 4-I8> From the Protractor Postulate and Theorem 4.3 it follows that there is a unique ray-coordinate system § such that cd VA = 0, < cd VD < 180,0 < cd v3< cd VD, andO < cd VC < cd VB. Let cd v8 = b, cdVC = ccdvB = d. We must show that if A AVB m Z CVD t then Z AVC S Z BVD. Since mLAVB = h- = fc mlCVD= d- c m£AVC = c m£BVD = d-b our problem amounts to proving that if b = d — c._ then c =z. d — b. Suppose then that b = d — c. Using the Addition Property' of Equal- ity, we may add c — h to both sides of this equation. The result is which simplifies to c = d — h, the desired conclusion. Notice tho similarity of Theorem 4.4 to Theorem 3.4. 158 Angles, Ray-Coordinates, and Polygon* Chapter 4 COEOLLAR 1 4,4, I If distinct rays VB and VCare between rays VA and V/3 and if Z AVC as Z BVD, then I AVB m Z CVD, Proof: If VB and V(? are between vX and VD, then vS and VB are between VA and VD. The corollary follows immediately from The- orem 4.4 by interchanging VB and VC, that is, by renaming ray VB as ray VC and renaming ray VC as ray VB. COROLLARY 4,4,2 If VA, VB. VC. VD are distinct coplanar rays such that A-V-D, B and C are on the same side of AD, and Z AVB at L CVD, then Z AVC =s Z BVD. Proof There is a unique ray-coordinate system in which cd VA = 0, cd VB = b, cd W= c, cd v3 - 180, b^ c, < h < 180, and 0<c< 180. Then mlAVB = b-0=b t ml AVC = c- = c, mLBVD- 180-6, mZCVD= 180 -c. Since Z AVB ss Z CVD, then b =s 180 - c. It follows that b + c = 180, c = 180 - fe, ml AVC = mlBVD, and ZAVC^= ABVD. EXERCISES 4.4 Exercises 1-10 refer to u ray-coordinate system in which the numbers as- signed to VA, V$. VC?, VD, V? ( and V?ure 0. 28, 47, 139, 263, and 319, respectively. In Exercises 2-9, compute the angle measures using the given ray-coordinates. 1. Draw a figure to illustrate the given situation. 2. mlBVC 6, mlFVE 3. m/ BVF 7, mZAVE 4. mlCVE 8. ml BVD 5. mlDVE 9. mlCVF 10. What can you say about rays VD and PF? 11. Let three distinct concurrent rays VA t VB, VA' such that VA = opp VA' be given. Prove that ml AVB + mlHVA' = 180, 4.4 Ray-Coordinates 159 12. la the situation of Exercise 11, explain why it is incorrect in our formal geometry to say that mLAVB + mLBVA' = mLAVA'. Kxercises 13-19 refer to a ray-coordinate system in which the numbers as- signed to VA, VB, VC, and VD are 0, b t c, and d, respectively, where 0<fr<e<ti< 180. In Kxercises 14-19, express the angle measures using the given ray-coordinates, 13. Draw a figure to illustrate the given situation, 14. mLAVC 17. m/CVD 15. mLBVD 18. m/AVD 16. mLBVC 19. mlAVB hi Exercises 20-20, the same situation as in Exercises 13-19 is given except that < b < c < 90 and 270 < d < 360. In Exercises 21-26, express the angle measures using the given ray-coordinates. 20. Draw a figure to illustrate the given situation. 21. mLAVC £4, m/CVD 22. mLBVD 25. mLAVD 23. mLBVC 26, mLAVB Exercises 27-38 refer to Figure 4-19. If the ray-coordinates of vX, vS, V6 t VD, VE, and VF are as .shown in the figure, name the theorem, definition, or postulate that justifies the statement in the exercise. U5G4 27. VD is between VS and VE. 33. LBVC == Z D VE 28. m Z CVE = ml CVD + 34. Z BVD m Z CVE m/DVE 29. V? is between V?and VB. 35. mZ AVE = 140 - = 140 30. mLBVF = ml BVC + 36. m I BVF = 180 - 40 = 140 m Z CVF 31. ml BVC = 70 - 40 = 30 37. LAVE BS Z BVF 32. mLDVE = 140 - 110 = 30 38. LAVB B LEW 160 Angles, Ray Coordfnales, and Polygons ■ Exercises 39 and 40 refer to Figure 4-20. Use Corollary 4.4.2. Chapter 4 i— * Figure 4-20 S 39. If ZM\W at / HVS, name an angle congruent to AMVR. 40. If /LSVNs* ZflVM> name an angle congruent to ZSVR. 41. In the figure below, L DBA S L CBE, What can you conclude about Z 1 and Z2? What are you assuming from the figure about the points A t B t C t 0, El About the points A, B, C in particular? About D and E in relation to AC? B D iL B 42. In the figure below, Z 1 — Z 2. What can you conclude about LBAE and Z CAD? Name the theorem you are using. What assumption are you making about the figure? 4.5 SOME PROPERTIES OF ANGLES Two distinct intersecting lines form four angles as suggested by Figure 4-21. Two angles such as a and c or h and d t which appear "op- posite" each other in the figure, are called vertical angles. Two angles such as a and b or h and c are called a linear pair of angles. We state these ideas more precisely in the following definitions. Figure 4-21 4.5 Some Properties of Angles 161 Definition 4.8 Two angles are called vertical angles if and only if their sides form two pairs of opposite rays. Definition 4. 7 Two angles are called a linear pair of angles if and only if they have one side in common and the other sides are opposite rays. Note that two angles are vertical angles if their union is the union of two distinct intersecting lines. Note also that two angles are a linear pair of angles if their union is the union of a line and a ray whose end- point lies on that line* THEOREM 4. 5 Vertical angles are congruent. Proof: Let AB and CD intersect at V to form two vertical angles A AVC and Z BVD as in Figure 4-22. Let a be the plane determined by A, V, and C Let % he the unique ray-coordinate system in a relative to V in which cd VA = and cd VC = mL AVC. (Which postulate tells us there is one and only one ray-coordinate system with these properties?) For convenience, let cd VB = b, cd VC a c t cd VD = d. b Figure 4-SS The rest of the proof follows from the definition of a ray-coordinate system. b = 180 and d = c + 180 Why? mlSVD =d- b=(c+ 180) - 180 = c = mlAVC L AVC ^ L BVD Definition 4,8 Two angles (distinct or not) are complemen- tary, and each is called a complement of the other if the sum of their measures is 90. Two angles (distinct or not) are sup- plementary, and each is called a supplement of the other if the sum of their measures is 180. 162 Angles, Ray-Coordinates, and Polygons Chapter 4 THEOREM 4.6 If two angles form a linear pair of angles, then they are supplementary angles. Proof: Let a linear pair of angles, Z AVB and Z BVC, in plane a as indicated in Figure 4-2-1 be given. Let § be the unique ray-coordinate system relative to Vin a such that cd VA = and cd VB = m L AVB. 4 • if- * ► Figure 4-23 A V C Then cd VJB < 180 and cd v2 = 180 ml AVB + mlBVC = (cd VB - 0) + (180 - cd VB) = 180. Therefore Z. AVB and ZBVC arc supplementary. Notice that Exercise 11 of Exercises 4.4 is Theorem 4.6. THEOREM 4. 7 Complements of congruent angles are congruent . Proof: Let Z A and £ B be two congruent angles, let ZC'bea com- plement of Z A, and let Z D be a complement of ZB. Then mlA + niLC- 90 -m/_8 + mlD = 90 mZA := roZB r»'ZA -h mZC = mZB + mLD mlC = mlD IC^ ID THEOREM 4.8 Supplements of congruent angles are congruent. Proof: Assigned as an exercise, A special ray associated with an angle is its midray. Here is a defini- nition for a midray. Deftnitfon 4.9 A ray is a midray of an angle if it is between the sides of the angle and forms with them two congruent angles. A midray of an angle is said to bisect the angle; it is sometimes called the bisector of the angle or, briefly, the angle bisector. 4.5 Some Properties of Angles 1 63 Tt appcais from Figure 4-24 that an angle should have one and only one midray, This suggests our next theorem. Figure i-M THEOREM 4. 9 Every angl e has a unique midray. Pmof: Let IAVC as in Figure 4-25 be given. Let § he the unique ray-coordinate system in which cd VA a and cd VC = c = ml AVC. Then there is a unique ray V£? such that cd VB s b m ^ Why? 2 ml AVE + m Z BVC = (fc - 0) 4- (<? - I) = c = mZAVC VB is between VA and VU. mZAVB = fc- 0= b mlBVC = c~ b=2b - h = h mlAVB = ml8VC Therefore VB is tetween VA and VC, and VB forms with V& and VC two congruent angles. Therefore VB is a midray of IAVC, We prove next that I AVC has only one midray. Suppose that VD is a midray of IAVC. Let cd VD = d, Then VD is between VA and V?. Why? < d < c by Theorem 4.3 mlAVD= d^0 = d mlDVC= v - d c — d = d, c = 2d, d= 4 = b VD = VB Why? Therefore the midray is unique. 164 Angles, Ray Coordinate*, and Polygons Chapter 4 EXERCISES 4,3 L In the figure several rays and their ray-coordinates are marked. Com- pute the measures of the following angles, (a) LAVB (b) LAVC (c) LAVH (d) LAVG (e) IAVF (f) LDVC (g) LDVB m LUVA 9 IDVH (1? LDVC :sn :i 2, In a ray-coordinate system the numbers assigned to VA. VB, VC, VD are 0, 30, 130, 180, respectively, (a) Draw a figure to illustrate the given situation. (b) Compute several angle measures and use them to prove that VB is between VA and VC. (c) Is VB between VA and VD? Justify your answer. (d) IsVA between V/sand VC? J ustify your answer. (e) Is VC" between VBand V0P Justify your answer. 3. In a ray-coordinate system die numbers assigned to VA and VB arc and 100, respectively. All of the rays and angles in this exercise are in the. plane determined by the points A, V, and B. (a) If m/.AVC = 50 and if B and C are on the same side of VA, find cd VC. (b) If m L AVC = 50 and if B and C are on opposite sides of VA, find cdvS. (c) If m Z BVD h 150 and if A and D are on the same side of VB, find cdvS. (d) If m / BVD b 150 and if A and D are on opposite sides of VB, find cd V5. (e) If cd VE = 200, us vS between VA and VE? (f) If cd VE = 200, is VA between V# and VE? 4.5 Some Properties of Angles 165 (g) If cd vS = 200, is Vj£ between V/l and VA? (h) If erf VF = 281, is VB between V# and V?? (i) If cd V? = 281, is VA* between vS and V?? (j) If cd VF = 281. is VF between VS and VA*? £k) Rod cd VK if VK b the midray of Z AVB. 4« Given a ray-coordinate system in which the numbers assigned to VA, VN, VD, VY are a, b, c, d, respectively, let VA\ VN\ v5\ v9* t bo the rays opposite to VA, VN, VD, VY, respectively. Assume that <£, a < b<ic < cf <C 1 SO. Derive formulas in terms of ft, b, c. d for the follow- ing measures. (a) mXAVN (e) mlDVA (b) mlAVY (f) mZJDVY (c) mLAVy (g) mLDVN (d) mLAVU (h) mtDVN 5, Referring to the figure, describe in your own words the following sets. (a) The union of VA and m (c) The union of Z AVB and Z BVC. (b) The union of VA and VB. (d) The union of Z AVB and / RVC. Referring to the figure, describe in your own words the following sets. (a) The intersection of VR and VF. \^p Xt (b) The intersection of ZiiVP and ZPVX (c) The intersection of Z H VP and I XVY. *_ (d) The intersection of LTVX and I XVP. 7. Let AM be the edge of a halfplane £JC that contains points C and D. As- sume that A, C, D arc noncoilincar and that B, C, D are noneollinear. Describe the following sets. (a) The intersection of 3C and AB. (b) The intersection of Z CAP and 3C. (e) The intersection of Z CAD and AB. (d) The intersectiOD of Z CAD and Z CBD. (e) The union of 3C and /LB. (f; The union of 3C and Z CAD, (g) The union of 0C and ZAGB. (h) The intersection of JC and ZACJ3. 166 Angles, Ray-CoordlnatM, and Potygorta Chapter 4 8. Let four coplanar rays VA, VB, VC", VD he given such that VB is be- tween VA and VC, VC is between VB and VD, and VD = opp vl as indicated in the figure. Complete the following proof that mLAVB + mLBVC + mlCVD- 180. Why? Why? Proof: m L BVD = w Z BVC + mL CVD mLAVB + mL BVD = 180 Therefore [TJ. 9. Let four coplanar rays VA, VB, VC, VD be given such that vS is be- tween VA and vS, v8 = opp V&, and m£AV€ > mLAVB as indi- cated in the figure. Justify each step in the following proof that mLAVB + ml BVC + mZCVD = ISO. . e B 4- B • M80 v Proof: Let § be the unique ray-coordinate system in which cd VA = cdvts mLAVB = b < 180. *— * — > Then B and C are on the same side of Ay, I ,et cd VC = c. Then < c < 180, where a = mLA VC. Then < b < c < 180. Therefore mZ AVB + m I BVC + mL CVD = (b - 0) + (o - b) + (180 - c) = 180, 10. In the figure AB and c3 intersect at O forming four angles, If tnLAOD = 133, find (a) ml CO A (b) mlBiX: (c) mZBOD 11. If LA and Z B are supplementary angles and i is a number such that mZA = 3* + 6 and mZ J? = &r + 12, find the measures of Z A and Z B. Check your results by finding the sum of these measures. 12. If LF and Z Q are complementary angles and y is a number such that mLF = y + 30 and mL{) s (/ — 30, find the measures of LF and Z (X Check your restilt by addition. 13. Twice the measure of an angle is 24 more than five times the measure of its supplement. Find the measure of the angle and check your result. 4.5 Some Properties of Angles 167 14. Find the measure of an angle if its measure is twice the measure of its complement, 15. Find the measure of an angle if its measure is one-half the measure of its complement. 16. If / A and Z B are both congruent and supplementary, find the mea- sure of each. 17. If Z A and Z B arc both congruent and complementary, find the mea- sure of each. 18. The figure shows tlirec eoplanar lines and six angles marked a„ b } c t d, x, and y. Complete the following state- ments, (There are several correct re- sponses for some items.] (a) aandi? are a [3 pair of angles, (h) a and b arc angles. (c) 6 and c are \J} angles. (d) If a sj x, then b s* [T]- 19. Given an angle L AVB, let 3C he the sot of all points that are on the same side of VB as A. that is, JC is the A-side of VB. Let X be the B-side of VA. Make a sketch showing JC by sliading with vertical h amines and X by shading with horizontal halflines. How is the interior of £AVB marked in your sketch? Is the interior of LAVB the intersection of 5C and X. or the union of 3C and X, or some other set related to 3Q and X? 20. In the proof of Theorem 4.7, justify each of the five equations and the congruence. 21. Prove Theorem 4.B. 22. In the figure below, AC is the midray of Z BAD and VS is the midray of LRVT. If lBAD=z IRVT, prove that ZJ?AC= ZRV5. £3. Speaking informally, does the result of Exercise 22 prove that "halves" of congruent angles are congruent? 24. Using Theorem 4.8 and the reflexive property of congruence for angles, write an alternate proof of Theorem 4.5. 25. In the figure at the right, Z2=s Z3. Prove that Z1^Z3 Z4^ Z5. 168 Angles, Ray-Coordinates, and Polygon* 4.6 INTERIORS OF ANGLES In Chapter 2 wc defined the interior of Z ABC as the intersection of two halfplanes, the C-side of A B and the A-side of BC. Another way to think of the interior of an angle is in terms of the rays between the sides of the angle. From our definition of between- ness for rays in Section 4.3, we know that if VD is a ray between the sides of LAVC y then every point of VD except V (that is, the hairline VD) lies on the A-sidc of VC and on the C-side of AV. Hence VD is con- tained in the interior of the angle. What we have shown for VD is true for every ray between VA and VC. For convenience let (R denote the union of the interiors of all rays between VA and VC and let $ denote the interior of / AVC. We have shown that (Ft lies in 3 t that is, that (Red. In Lemma 4. 1 0. 1 we shall show that if a point is in §, it is also in (ft. Then we shall have (R C d and £ C (R, and hence $ =s (R, which we state formally in Theorem 4.10. LEMMA 4.10.1 If a point is in the interior of an angle, then it is an interior point of a ray between the sides of that angle. Proof: Let D be a point in the interior of Z.AVC, and let a be the plane containing the points A, V, C, D. Then it follows from the defini- tion of the interior of an an^le that (1) A and D arc on the same side of VC and (2) D and C are on the same side of VA. We shall show diat VD — ? > is between VA and VC y and hence that D is an interior point of a ray between the sides of L AVC, Let § and §' be ray- coordinate systems in a relative to V with co- ordinates as indicated in the table and such that <C c <C 180. < a' < 180. Since < c < 180 and since C and D are on the same side of VA, it follows that < d < 180. Since < d < 180 and since A and D are on the same side of VC, it follows that < & < 180. S 3L cdVA a' cdVD d d' ud VC? c 4.6 Interiors ot Angles 169 Suppose that c = d. Then d' = (Why?), D lies on VC t and D is not in the interior of Z.AVC. Since this contradicts the hypothesis of the theorem, it follows that c^d. Then < d < c < 180 or < c < d < 180. Suppose that < c < d < 180. Then it follows from Theorem 4.3 that VC is between VA and VD and from Definition 4.3 that A and D are on opposite sides of VC. However, A and D are on the same side of VC, (See (1) in proof.) Since this con- tradicts the Plane Separation Postulate, il follows that < d < c < 180. Using Theorem 4*3 and Definition 4.3 again, we deduce that YD is between VA and VC and hence that D is an interior point of a ray be- tween the sides of £ AVC, We have proved the following theorem, THEOREM 4.10 The interior of an angle is the union of the in- teriors of all rays between the sides of the angle. Another way to consider the interior of an angle is in terms of the segments whose endpoints he on the sides of the angle. (Sec figure 4-26.) Figure 4-26 Given Z ABC, let D be any point on BA except B and let E be any point on BC except B< Let P be any interior point of the segment DE. In other words, P is any point between D and E. Since all of DE except E lies on the A-side of BC and all of DE except D lies on the C-side of AB t it follows that Plies in the interior of A ABC. Thus we have the following theorem. THEOREM 4.11 If AB is a segment joining an interior point of one side of an angle to an interior point of the other side, then the interior of AB is contained in the interior of the angle. 170 Angles, Ray-Coordinates, and Polygons Chapter 4 4.7 ADJACENT ANGLES AND PERPENDICULARITY A linear pair of angles is a special case of a pair of coplanar angles having one side in common and interiors which do not intersect. (See Figure 4-27.) It is convenient to introduce a special word for angles with th is property. 4 Figure 4-27 Definition 4.10 Two coplanar angles are adjacent angles if they have one side in common and the intersection of their interiors is empty. "fl Figmc 4-2S In Figure 4-28, L AVE and Z B VCare adjacent angles and Z AVE and Z AVC are adjacent angles. Although Z DWE and Z EWF arc adjacent angles, note that Z DWE and Z DWFaxe not adjacent angles. All of you have a background of experience with right angles, per- pendicular lines, acute angles, and obtuse angles. Following are the formal definitions for these terms. Definition 4.11 An angle whose measure is 90 is a right angle. An angle whose measure is less than 90 is an acute angle. An angle whose measure is greater than 90 is an obtuse angle, (See Figure 4-29.) Sight EUjgle Figure 4-2M AouU- nriKle Oi'LusL' im^Uti 4.7 Perpendicularity 171 THEOREM 4. 12 If the two angles in a linear pair are congruent, they are right angles. Frew?/: (See Figure 4-30.) Let the measure of each angle in the linear pair be r. It follows from Theorem 4.6 that r + r = ISO, Therefore r = 90 and each of the angles is a right angle. Figure 4-30 THEOREM 4.13 Any two right angles are congruent. Proof; Every right angle has a measure of 90. Hence all right angles have the same measure and hence all right angles are congruent to each oilier. Definition 4.12 If the union of two intersecting lines con- tains a right angle, then the lines are perriendicular. If Z A VB is a right angle, then AV and VB are perpendicular (See Figure 4-31.) mcs. • A ~V A B' >A' Figure 4-31 Figure 432 It b easy to show that two perpendicular lines form four right angles. Let A A' and BB' be perpendicular lines which intersect at V as indicated in Figure 4-32 and let L AVB be a right angle. Then LA'VW is a right angle since LA'VB' and Z AVB arc vertical angles and vertical angles are congruent. Also, L AVB r is a right angle since I AVB' and A AVB form a linear pair. Then ml AVB = 90, mLAVB + ml AVB' = 180, and therefore ml AVE' = 90, Why is Z A' VB a right angle? 172 Angles, Ray-Coordinates, and Polygons Chapter 4 When are two rays perpendicular? It seems natural enough to say that two rays are perpendicular if the lines which contain them are per- pendicular. We extend this idea to any combination of segments, rays, and lines in the following definition. Definition 4. 13 Two sets, each of which is a segment, a ray, or a line, and which determine two perpendicular lines are called perpendicular sets, and each is said to be perpendicu- lar to the other. Notation. We write © _L (B to mean that <2 and (B arc perpendicular sets, Wc may read <$. _L (B as "'d is perpendicular to (&' f Note that if a 1 (B, then OS 1 4 If d and (B are perpendicular sets, then & is contained in some line I, (B is contained in some line m, and I JL m. Since perpendicular lines are distinct intersecting lines, it Follows that if Q, _L (B. then d and (B have at most one point in common. As indicated in Figure 4-33, the intersection of two perpendicular sets is a set consisting of at most one point. A ray porpoodieular to a line; intersection is empty ray perpendicular tn n ray; interaeclion \s hmi;!v Figure 4-33 4.7 Perpendicul»rtty 173 In a figure two perpendicular sets may be marked with a little "square comer" as illustrated in Figure 4-34. 3. Figured A typical exercise in informal geometry involves constructing a per- pendicular to a line at a point on it. Our formal geometry is sufficiently developed now so that we can prove that such perpendiculars exist. THEOREM 4. 14 For each point on a fine in a plane, there is one and only one line which lies in the given plane, contains the given point, and is perpendicular to the given line. Proof: See Figure 4-35. Let P be a point on line I in plane a. Let 0C he one of the two halfplanes in a with edge I. There exists a point A different from P on /, Why? B X • R Figure 4-io Let R be any point in 3C. Let § be the unique ray-coordinate system in a relative to P in which cd PA = and cdPR = m/ APR. Let PR be the unique ray with cd T3 = 90. LAPS is a right angle andPB± PA. Why? Therefore there is at least one line in a through P perpendicular to t. Let m be any line in a through P and perpendicular to I Then m U /contains four right angles. One of these right angles is the union of FA and a ray, say PC, such that PC? C 3C. m£APC = 90 0<cdpd< 180 cd PC = 90 PC = PB Why? m = PE Why? Hence there is only one line in a through P and perpendicular to L 174 Angles, Ray-Coordinates, and Polygon* Chapter 4 In connection with Theorem 4.14 there is an interesting question to consider. If / is a line and F is a point not on I, is there one and only one line through P that is perpendicular to T? (See Figure 4-36.) Maybe there is no line m such that m L i. Maybe there is Just one such line m. Maybe there are several* We shall prove later that in our formal ge- ometry there is one and only one line m through P and perpendicular to I. The point in which m intersects I is called the foot of this perpen- dicular from P to I. p +i Figure 4*36 It is important to emphasize that all lines considered in Theorem 4.14 lie in one plane. Tf we remove this restriction, it seems reasonable that there are many lines perpendicular to a given line at a given point on it, as suggested in Figure 4-37. This idea comes up again in a later chapter. Figure +a? EXERCISES 4,7 Let a ray-coordinate system § in a plane n relative tea point Vbe such that the ray-coordinates of VA, VB, VCarc a, b> c, respectively- In Exercises 1-5, given the values of a, 6, G, determine whether or not B is in the interior of IAVC. I a = t b = 90 t c = 175 2. a = 0, b a 90, c = 185 3. a = 0,b = 270, c = 185 4. a = 0, h - 270, c = 175 5. a = 90, h = 260, c = 4.7 Perpendicularity 175 In Exercises 6-16, A t B, Care three noncollinear points and D is an interior point of BC. 6. Is D an element of the interior of Z J9AC? 7. Is D an element of the interior of Z BCA? 8. Is D an element of the interior of / CAB? 9. Is J? an element of the interior of Z CAU? 10. Is C an element of the interior of Z DAB? 11. Is AB a subset of I ABC? 12. Is AB a subset of I ABC? 1 3. Is the interior of A~fi a subset of Z ABC ? 14. Is SUa subset of /.ABC? 15. Is AD a subset of die union of Z.BAC and its interior? 16. Answer each question under "reason" with a definition, a postulate, or a theorem in the following proof that mZ DAC + mLDAB = m£BAC. atcment (a) D is an interior point of BC. (b) D is in the interior of ABAC. (c) AD is between A~B and AC. (d) mLDAC + mLDAB = mLBAC (a) Given (b) Why? (e) Why? (d) Why? ■ In Kxereises 1 7-22, state which of the following descriptions, (a) to (e), cor- rectly identifies the perpendicular sets in the given diagrams. (a) Two perpendicular rays, (b) Two perpendicular segments, (c) A line and a ray perpendicular to each other. (d) A ray and a segment perpendicular to each other, (e) A line and a segment perpendicular to each other. 17. <- + 20. 21. 22. t \ 176 Angles, Ray-Coordi nates, and Polygon* Chapter 4 ■ In Exercises 23-27, supply the missing word so that each sentence is Inie. 23. If the angles in a linear pair of angles are congruent to each other, I hen each of them is a \T\ angle. 24. If the measure of an angle is less than 90. then it is an \T\ angle, 25. If the measure of an angle is greater than 90. then it is an [T| angle. 20. If one angle of a linear pair of angles is acute, then the other one is [TJ. 27- If A ABC is a right angle, then BA and S?are [?] lines. ■ Let a set of coplanar rays be given as in Figure 4-38 such that m Z VAX = 130, m/MVR = 65, m£SVM = 30, VT is the midray of ZflV?>, VS is between VM and VR t VR is between VS and VD. Use this information to find the measure of the given angle in Exercises 28-32. 2& IRVS 29. ZDVR 30. ZDVT 31. ZDVS 32. ZrVTVf Figure 4-3S ■ In Figure 4*39, A3, CD f EF are coplanar lines that intersect at V and mLAVC = 32.3 and m£AVF = 151.7. Use this information to find the measure of the given angle in Exercises 33-37. 33. ZCVF 34. LFVB 35. LBVE 36. AEVA 37. AEVC Figure 4.30 ■ In Exercises 38-42, Z AVR in plane a and a ray-coordinate system in « such that cd VA = 270 arc given. In each exercise, supply the two numbers that are as close together as possible and that make the resulting statement true, 38. If A AVB is obtuse and cd vS < 90, then cd VB is a number between (T|aiid{¥J. 39. If Z AVB is acute and cd VS < 270, then cd VB is a number between {?] and [T]. 40. If Z AV.fi is obtuse and cd vB > 90, then cd VB is a number between \T\ and [7|. 41. If I AVB is acute and cd v3 > 270 } then cd V$ is a number betwoen [T]and[7]. 42. If/ AVB is a right angle, then cd V$ is [7] or [|| i 4.8 Polygons 177 43. Copy and complete the proof of the following theorem. Draw a figure to help you understand the result. THEOREM U I A VB, Z BVC, and L CVA are three coplanar angles such that the interiors of no two of them intersect, then the sum of the measures of these three angles is 360. Proof: In plane A VB there is a unique ray-coordinate system § relative to Vin which cdvX = and cd VB = ml AVB. From the [T] Postulate it follows that m Z AVB is less than \J] t hence that cd VB is less than [T|. By JTj the interiors of A. AVB and /.AVC do not intersect. Therefore C and B lie on [T] (the same side, opposite sides) of AV. Therefore the ray-coordinate of VC, call it c> is [T] than 180. Since the interiors of Z BVC and Z AVC do not intersect, it follows that c is less than h + [r\. Then mLAVB= h -0 = b mlBVC = \2\ mLCVA = fT] mZAVB + mlBVC + mLCVA = b + (c - b) + = 30Q 44. In the proof of Theorem 4J4 there is a statement that m J ! is the union of four rigjit angles. One of these angles is described in terms of PA arid a ray PC such that PC C 3C. Describe the other three angles in a similar way, 4.8 POLYGONS As stated before, triangles are three- sided polygons and quadri- laterals are four-sided polygons. A polygon is a plane figure with n ver- tices and n sides, where n is an integer greater than or equal to 3, The figures shown in Figure 4-40 are polygons, but the figures shown in Figure 4-41 are not. Ffpir»4-4J 178 Angles, Ray-Coordinates, and Pelyfont Our formal definition is as follows, Chapter 4 Definition 4.14 Let n be any integer greater than or equal to 3, Let Pi, P 2 . . . . , P n u P* he n distinct coplanar points such that the n segments P,P 2l P^, P^-iA, Jyi have the following properties: 1. No two of these segments intersect except at their endpoints. 2. Xo two of these segments with a common endpoint arc collinear. Then the union of these n segments is a polygon. Each of the n given points is a vertex of the polygon. Each of the n seg- ments is a side of the polygon. Tf n — 3, the definition of a polygon yields a triangle; if n — 4, it yields a quadrilateral. Sometimes a polygon with n sides is called an n-gon. For example, a polygon with 13 sides is a 13-gon. The following list gives the names commonly used for the polygons having the num- ber of sides indicated. You should learn these names. • umber of Sides N ante 10 12 'triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Decagon Dodecagon Notation. The polygon whose vertices are Fi, F& . . . a P„ and whose sides are TJ^ t Pp^, .... P« ^.P^is called the polygon PjP 2 . . . P„. Definition 4. 15 Two vertices of a polygon that are end- points of the same side are called consecutive vertices. Two sides of a polygon that have a common endpoint arc called consecutive sides. A diagonal of a polygon is a segment whose endpoint s are vertices, but not consecutive vertices, of the polygon. 4.8 Polygons 179 Most of our work with polygons is restricted to polygons like those ill Figure 4-42, Figure 4-4£ We usually exclude polygons like those in Figure 4-43 because they are not convex polygons. Figure 4-43 The polygons that interest us are the convex polygons, The follow- ing is our formal definition. Definition 4. 16 A polygon is a convex polygon if and only if each of its sides lies on the edge of a half plane which con- tains all of the polygon except that one side. You in ust be careful not to confuse the idea of a convex polygon with that of a convex set as defined in Section 2.5. A convex polygon is not a convex set of points, although the union of a convex polygon and its interior is (sec Definition 4,17). Figure 4-44 illustrates Definition 4-16, It shows the halfpkncs as- sociated with the sides as required in the definition. 180 Angles, Ray-Coordinates, and Polygon* Chapter 4 Figure 4-45 shows a quadrilateral ABCD that is not convex. Neither the D-side of BC nor the A-side of BC contains all points of the quadri- lateral not on BC. Draw a figure to convince yourself thai every triangle is a convex polygon. Ffgurc+-*5 Definition 4.17 The interior of a convex polygon is the in- tersection of all of the Imlfplanes, each of which has a side of the polygon on its edge and each of which contains all of the polygon except that side. Figure 4-46 illustrates this definition for a convex quadrilateral ABCD. 3C = C-side of A~§ 6 = D-side of BC? g = A-side of cB DC = B-side of Ha Interior of ABCD = 3C (1 £ O $ PI 3C Definition 4, 18 An angle determined by two consecutive sides of a convex polygon is called an angle of the polygon. Two angles of a polygon are called consecutive angles of the polygon if their vertices are consecutive vertices of the polygon. For a triangle, any two of its vertices are consecutive vertices, any two of its sides are consecutive sides, and any two of its angles are con- secutive angles. For a quadrilateral we use the word opposite when consecutive is not applicable as in the following definition. Definition 4.19 If two sides (or vertices, or angles) of a quadrilateral arc not consecutive sides (or vertices, or angles), then they are opposite sides {or vertices, or angles) and each is said to be opposite the other. 4.9 Dihedral Angles 181 Figure 4-47 illustrates this definition. For this example we have A and C are opposite vertices. B and D arc opposite vertices, AB and CO are opposite sides, BC and 73^ are opposite sides, LA and / C are opposite angles, L B and L D are opposite angles. Figur* 4-47 4.9 DIHEDRAL ANGLES Angles are plane figures. Every angle is a subset of a plane. Closely related to the idea of an angle is the idea of a dihedral angle. Sometimes we say plane angle when we want to emphasize that an angle is not a dihedral angle. A plane angle is the union of two noneollinear rays hav- ing the same endpoint. A dihedral angle is formed by two halfplanes and a line. Here is the formal definition. Definition 4*20 If two noncoplanar halfplanes have the same edge, then the union of these halfplanes and the line which is their common edge is a dihedral angle, The union of this common edge and either one of these two halfplanes is a Face of the dihedral angle. The common edge is the edge of the dihedral angle. A dihedral angle is suggested by Figure 4-48. In this diagram, B and C are poinfe on the edge, A is a point in one face but not on the edge, D is a point in the other face but not on the edge. A suitable name For this dihedral angle is A-BC-D or LA-BC~T>. The letters at both ends of this symbol are names of two points, one in each face but not on the edge; the two letters in the middle are names of distinct points on the edge. Am Figure 4-48 182 Angles, Ray-Coordinates, and Polygon* Chaptor 4 The same way two intersecting lines form four angles, two inter- secting planes form four dihedral angles as indicated in Figure 4-49. LAVE LDVC LCVD Z.DVA Figure 4-49 Definition 4,21 Two dihedral angles, such as A-UV-B and D-IJV-C in Figure 4-49, which have a common edge ami whose union is the union of the two intersecting planes are vertical dihedral angles. EXERCISES 4.9 I, Copy and complete the following definition of a pentagon: Let F, Q, R t S, T be five df&ttoct eopbmar points such that the five seg- ments [T|, EH, EH* S [D have the following properties. (a)[lJ mm Then the [T] of these five segments is a pentagon. In Exercises 2-7, one of the sides of hexagon ABCDEF is given. In each exercise, name two sides which arc consecutive with the given side. 5. DE 2. AB 3. BC 4. CD 6. IF 7. FA 4.9 Dihedral Angles 183 8. Copy and complete the following proof that if ABCD is a convex quad- rilateral, then the interior of AC is contained in the interior of the quadrilateral. Proof; Let if denote t he interior of AU a nd let $ denote the i nterior of the quadrilateral. U C C-side of AB C C-sidc of AD C A-side of BC C A-side of CD* But g is the intersection of [Tj. Therefore $ gj. 9. (a) Draw a convex pentagon and all of its diagonals, (b) How many diagonals does a convex pentagon haver (c) Does the interior of each diagonal of a convex pentagon lie in the interior of the pentagon? 10. Draw a pentagon which is not convex. Label it and explain why it is not convex. 11. Draw all of the diagonals of the pentagon you drew in Exercise 10- I low many diagonals are there? Figure 4-50 shows a labeled cube. In Exercises 12-16, a quadrilateral whose vertices arc vertices of this cube is given. la each exercise, write a suitable name, using four of the letters from the figure, for a dihedral angle contain- ing the given quadrilateral. 12. ABCD 13. ABFE 14. BCGF 15, CGHD 16. I'Mlll B Plgvn 450 Exercises 17-21 also refer to Figure 4-50. Base your answers on examining the figure. In some cases the formal geometry that we have developed up to this point is inadequate to prove the correctness of these answers. 17. Is A-BF-H a dihedral angle? 18. Is A-BF-C a dihedral angle? 19. Is A-BF-E a dihedral angle? 20. Is A-BF-D a dihedral angle? 2L Is G-HE-C a dihedral angle? 184 Angles, Ray-Coordinales, and Polygon* Chapter 4 In Exercises 22 and 23, a dihedral angle A-BC-D* a line I in plane ABC through A and not intersecting BC, and a line m in plane DBC through D and not intersecting BC arc given. 22, Draw a figure to illustrate this situation. 23. Explain why I and m do not intersect. 24* Figure 4-51 shows two planes, a and & intersecting in line AB. How many dihedral an^es arc formed by these two intersecting planes? Name them. 25. I>efine the interior and exterior of a dihedral angle. 26. Define adjacent dihedral angles. 27. Name a pair of adjacent dihe- dral angles in Figure 4-51. 28. Name a pair of vertical dihedral angles in Figure 4-51 . Figure 4-51 CHAPTER SUMMARY The central theme of this chapter is properties of angles. We intro- duced three new postulates; The ANGLE MEASURE EXISTENCE POS- TULATE, the ANGLE MEASURE ADDITION POSTULATE, and the PROTRACTOR POSTULATE. The definitions include the following. CONGRUENT ANGLES MEASURE OF AN ANGLE BETWEENNESS FOR RAYS RAY-COORDINATE SYSTEM VERTICAL ANGLES LINEAR PAIR OF ANGLES RIGHT ANGLE OBTUSE ANGLE ACUTE ANGUS COMPLEMENTARY ANCLES SUPPLEMENTARY ANGLES MIDRAY OF AN ANGLE ANGLE BISECTOR ADJACENT ANGLES PERPENDICULAR SEGMENTS, RAYS, AND LINES POLYGON CONVEX POLYGON DIHEDRAL ANGLE Review Exarclses 185 This list of terms should remind you of many of the ideas and theorems of tliis chapter. Ray-coordinates have been used extensively in developing the ideas and in proving the theorems that are included Theorems 42 (ANCLE CONSTRUCTION THEOREM), 4,4 (ANGLE MEASURE AD- DITION THEOREM). 4.5, 4.6, 4.7, and 4.8 are important for the work that you will do in Chapter 5. Be sure that you know die statements of these theorems. REVIEW EXERCISES In Exercises 1-10, write definitions for the following terms. 1. Congruent angles 2. Vertical angles 3. Adjacent angles 4. linear pair of angles 5. Complementary angles 6. Supplementary angles 7. Right angle 8. Acute angle 9. Obtuse angle 10. Dihedral angle In Exercises 1 1 -20. a, b, c are the ray-coordinates of VA, VB, V T C, respec- tively. In each exercise, determine if one of the three rays is between the other two. If one is, name it. If none is, write "none,* 1 11. a = 0, b = 10, c = 170 12. a = 0, b = 10, c = 180 13. a = t b= 10, c = 190 14. a = 0, b = 10, c - 200 15. o = 350, h= 50, c= 110 16. a = 270. b = 180, c = 135 17. a = 270. b m a c = 135 IS. a = 180, h = 90, a = 20 19. a = 135, b = 315, c = 20. a. = % b m 300, c = 100 186 Angles, Ray-Coordinates, and Polygon! Chapter 4 ■ In Exercises 21-30, a and b sire the ray-coordinates of VA and VB, respec- tively. In each exercise, compute the measure of / AVB. 2L a = 38, b = 106 22. a = 300. fo = 150 23. a = 300, b = 100 24. a = 359, h = 1 23. o ^ 270, h = 100 26. a = 38, & = 50 27. a = 198 s b = 28. a = 15, fc = 300 29. a = 6, b = 40 30. a = 315, b = 345 ■ In Exercises 31-40. cdVA = cdvf= 150 tfd W? a 10 cd VK - 200 afVC?=20 cdv£ = 2S0 ar\S = 30 crfVM=300 crfv£=40 «/V>f=305 cd V?=50 cdVP = 310 erf v2 = 75 erf VQ = 315 cd VH = 90 erf V7* = 325 erf V?= 100 crfVS = 330 In each exercise, determine whether or not the given statement is true. 31. LBVffm IJVL 32. Z7VH = LDVE 33. Z SVB = Z UVF :I4, Z B VE and Z DV7J are supplementary angles. 35. Z CVE and Z SV7 are supplementary angles. 36. Z.KVL and LAVE are complementary angles. 37. W? is the midray of L DVJ. 38. V0 is the midray of L MVP 39. V^ k the midray of Z KVE, 40. Z SVil is an acute angle, 41. Explain how betweenness for rays is related to betvveenness for ray- coordinates. Review Extremes 187 42. Explain bow betweenncss for rays is related to the addition of angle measures. 43. Ii LA and Z B are complementary angles and if mZA = X and mlB= 3* + 10, findr. 44. If Z T and Z W are supplementary angles and if mLT = X and mZW=3*+ 10, find x. 45. Use ray-coordinates to prove that vertical angles are congruent. 46. Use ray-coordinates to prove that the angles of a linear ]>air are supplementary, 47. If A-V-A' r B-V-ff, cdVA = 143, cd vS<cdVA, and tX' I «"#\ find ed VB, cd W, cd VB'. 4 S. 1 f A BCD is a convex quadrilateral and £ is a point not in the plane of tiie quadrilateral, which of the fallowing are names of dihedral angles: A-BC-B, A-BC-E, D-BC-E, E-AD-C? 49. Explain why a convex polygon is not a convex set. 50. Explain why a dihedral angle is not a convex set. Chapter CtHirtvsy of Quebec Government TouriH Office Congruence of Triangles 5.1 INTRODUCTION Suppose you placed a sheet of carbon paper between two sheets of paper and drew a picture of a triangle on the top sheet Will the carbon copy of the triangle that appears on the second sheet be the same size and shape as the one you have drawn on the top sheet? The idea of two physical objects being carbon copies of one another is what we have in mind when we say that the two objects have the same size and shape. Obviously, we cannot take a physical object such as a house and, using carbon paper, make a carlxm copy of it We can, however, take the same set of blueprints that were used in constructing one house and construct another house that is exactly like it, diat is, having the same size and shape as the first one. In this chapter we are concerned witii the "size and shape" of ge- ometrical objects such as segments, angles, and triangles. These geo- metrical objects are not the physical objects that we draw on our paper. They are the mathematical objects which exist in our minds and whose properties have been described in our postulates, definitions, and the- orems. The mathematical concept corresponding to "same size and shape" is congruence. We, have already defined congruence of seg- ments in Chapter 3 and congruence of angjes in Chapter 4. Since all segments have the same shape, we say that two segments are congruent 190 Congruence of Triangle* Chapter 5 if they have the same size, that is, if the measures of their lengths are equal. Similarly, two angles have the same size and shape or are con- gruent if their measures are equal 5.2 CONGRUENCE OF TRIANGLES In order to arrive at a definition of congruence for geometrical tri- angles, let us consider again two physical triangles, one of them a car- bon copy of the other. Consider first a triangle that has sides of three different lengths, a scalene triangle. The definition we will obtain ap- plies, however, to all triangles regardless of their shapes. Figure 5-1 shows two scalene triangles, A ABC and its carbon copy. AA'B'C If you arc to make A ABC "fit" on AA'B'C, you must match up the vertices of the two triangles according to the scheme in Figure 5-1: A <— * A', B «— ► B', C <— + C. This matching is called a one-to-one correspondence between the two sets of vertices. The correspondence between the vertices of the two triangles also gives a correspondence between the sides of the triangle. A75 ISC A'W, If a correspondence between the vertices of two triangles is such that the corresponding angles and corresponding sides of the two tri- angles are congruent then the correspondence is called a congruence between the two triangles. The correspondence we have been discuss- ing is a congruence between A ABC and AA'&C. On the other hand, 5.2 Congruence of Triangles 191 it is possible to write a correspondence between the vertices of the two triangles that is not a congruence. For example, the correspondence, A « — * B\ B +-* C t C ^^ A\ is not a congruence between the two triangles shown in Figure 5-1 be- cause, by this matching of vertices, it is not possible to make A ABC coincide with &A'WC. Write four more correspondences between the vertices of A ABC and AA'FC such that no two are the same and such that none is a congruence between the two triangles. It is convenient to write a correspondence, such as A < — ► A\ B <— » W, C<— » C, on one Une as ABC ± — * A'B'C When this notation is used, it is un- derstood that the first letter on the left corresponds to the first letter on the right of the double arrow, the second corresponds to the second, and the third corresponds to the third as shown below. tf BC <— ► A'B'C V .. |» 1 Note that there arc several ways of writing this same correspond- ence, For example, both BAC 4 — * B'A'C and CBA « — » CB'A' name the same correspondence as ABC * — * A' WO. The foregoing discussion about corresponding vertices of physical triangles can be made to apply equally as well to abstract geometrical triangles. Thus, if ABC i — > DEF is a correspondence between the vertices of any two geometric triangles, this correspondence provides us with three pairs of points. We are interested primarily in these points as vertices of the angles of the triangles and as endpoints of the sides of the triangles. In connection with this correspondence we speak of six pairs of corresponding parts. Three of these six pairs are pairs of angles: LA *-^ ID, L B < — > L E, IC ^— £R The other three pairs are pairs of sides: AB < — > Z5E, BC < — ► Iff, au ^ ra 192 Congruence of Triangles Chapter 5 We are now ready lo state the definition of a congruence between two triangles. Definition 5.1 Two triangles (not necessarily distinct) are congruent if and only if there exists a one-to-one correspond- ence between their vertices in which the corresponding parts are congruent. Such a one-to-one correspondence between the vertices of two congruent triangles is called a congruence. If ABC < — > DEF is a congruence between A ABC and ADEF, then we write A ABC at A DEF and note that the following six state- ments are true: I A aZD, AB^DE, in?- lE r BC=zEF, ACstLF, AC = DF. We can also say that, for A ABC and A DEF, if all of these six state- ments are true, then ABC * — * DEF is a congruence, or what means the same thing, AABC as ADEF, In view of the definition of con- gruent triangles, we sometimes say that "corresponding parts of con- gruent triangles arc congruent," If AABC == ADEF, explain why each of the following six equa- tions is true: m£A = m£D t AB = DE, m£B = m£E, BC = EF, tn£C=mlF t AC = DF. Note that if AABC and ADEF arc distinct triangles and if ABC < — > DEF is a congruence, then it is correct to write AABC ss ADEF t but that it is incorrect to write AABC = A DEE The state- ment "AABC ==; A DEF" is a short way of saying "ABC « — * DEF is a congruence;" it is a statement about a one-to-one correspondence between the vertices of two triangles. The statement "AABC = ADEF" is a statement that two sets arc equal; it means that AABC and ADEF are names for the same triangle. Note also that if AABC ss A DEF t then ABAC == A EDF, For AABC B ADEF means that ABC * — > DEF is a congruence; ABACS A EDF means that BAC < — * EDF is a congruence; and ABC 4 — > DEF and BAC < — > EDF are two ways of describing the same one-to-one correspondence between vertices. On the other hand, if ABC i — > DEFh a congruence but ABC < — > FED is not a con- 5.2 Congruence of Triangles 193 gruence between the two triangles, then it is incorrect to write AABCm A FED. In drawing figures it is convenient to label congruent angles and congruent sides of triangles with the same number of marks as shown in Figure 5-2. c Figure 5-2 of the six congruences indicated in the figure are LA^LF, IB^IE, /.C^ID. Name the three pairs of congruent sides indicated in the figure. The six congruences tell us ABC < — > FED and A ABC m AFED. It is also helpful to label the side that Is opposite a given angle in a triangle with the same number of mark* used in labeling the angle, as has been done in Figure 5-2. For example, in AABC t L C and side^fB (which is opposite LC) both have the same number of marks. EXERCISES 5.2 Exercises 1-6 refer to Figure 5-3. D Figure W EOF is not a, congruence. A V3 * * VI 1. Is AABC ^ ADEF? Why? 2. Explain why the correspondence ABC * — ► 3. Is it correct to write AABC ss AEDF? 4. Is it correct to write ABC A = AEFD? Why? 5. Write four more Statements of congruence between the two triangles, each of which foikiws immediately from the fact that ABC * — » DEF is a congruence. 6. Write three correspondences between the vertices of the two triangles such that each is not a congruence, 194 Congruence of Triangles Chapter 5 7. The following figure shows two scalene triangles with corresponding congruent sides and angles marked alike. Copy and complete the following correspondences so that the re- sulting statements are true, LKP * — * \T\ is a congruence, LPK « — > \J} is a congruence, KLP < — * [T] is a congruence. KPL 4 — » [TJ is a congruence. PKL * — > is a congruence, PLK <- — * [f] is a congruence. & In Exercise 7, why is the correspondence LKP * — * KST not a congruence? 9. In the figure. AA£Ca APQB. Copy and complete the following statements by supplying the missing symbols. The correspondence A[T| * — * \]}QR is a congruence. Lh*k LP AE^^l 10. In the figure, AABCs A DBG. List the six pairs of corresponding, congruent parts of these two triangles. 5,2 Congruenco of Triangles 195 1 1 . In Exercise 10, if £ ABC and Z DBC are distinct coplanar acute angles, does my B? bisect I ABD? Wiry? 12. In Exercise 10 (with the figure appropriately modified), if L ABC and L DBC are coplanar obtuse angles, does ray BC bisect Z AHDP Why? 13. The figure below shows eight triangles. If two triangles look congruent, assume that they really arc congruent. Write congruences between such congruent pairs of triangles. 14. Without using a figure, list the six pairs of corresponding, congruent parts for the triangle congruence AEFK =s AABT. 15. Without using a figure, list the six pair* of corresponding, congruent parts for the triangle congruence ARPSss ALSR 196 Congruence of Triangles Chapter 5 16. For A ABC and A CAS, it is true that AH - $C, AC. - SA, #C as CA, Z A as IS, Z C as Z A, and Z A as Z G. Write a statement of con- gruence between these two triangles. 17. If &DEFis a scalene triangle* prove that the statement ADEFsz ADFE is false, 18. If A AST as ASIA, what special property does A A ST have? Draw a suitable figure for A AST". 19. If A/Jfitf as A MEN, what special property does ALMA r have? Draw a suitable figure for ALMN, In Exercises 20 and 21, complete the proof of the following theorem. THEOREM Congruence for triangles is reflexive, symmetric, and transitive. Proof: Let A ABC be any triangle. Then IA as ZA» Z B £* Z A t and ZCas ZC by the reflexive property of congruence for angles. Also AA=?AB, WlrsBCt and AC as AC by the reflexive property of congruence for segments. Therefore A ABC as AAAC by the definition of congruent triangles. Therefore congruence for triangles is reflexive. 20. Prove that congruence for triangles is symmetric 21. Prove that congruence for triangles is transitive. 22. In the figure at the right, CD 1 AB, AC as BC, AD = BD t LACD as Z BCD, m£A = SO, mZB= 00. Prove that A ACD as A BCD, 23. In the figure at the right, Af is the midpoint of §7, Mi s the midpoint of 7ZF., KS as EJ> mZS=_r»ZJ, W J KE f IE ± KE. Prove that A SKM as A JEM. 5.3 "If-Then" Statements 197 5.3 "IF THEN" STATEMENTS AND THEJR CONVERSES Many of our definitions, postulates, and theorems have been stated in the "if-then" form. They have been statements of the type "If p, then tf," where p and q are statements, (Remember that a statement is a sentence which is either true or false.) In other instances we have used the phrase "if and only if ** in the statement of some of our defi- nitions, postulates, and theorems. As tin example of the use of the phrase "if and only if consider die definition of congruent segments given in Chapter 3: Two segments are congruent if and only if they have the same length. This statement is a conjunction of the following two statements: L Two segments are congruent if they have the same length. 2. TVo segments arc congruent only if they have the same length. Statements 1 and 2 can be restated in the "if-then 1 * form as follows: 3. It two segments have the same length, then they are congruent. 4. If two segments are congruent, then they have the same length. We note two important things about these last two statements. (a) Each is written in the "if-then" form. The if-clause of each statement is the then-clause of the other. (This also means that the then-clause of each is the if-clause of the other.) (b) Both statements may be used in proofs. For example, if we know AB as n? and want to establish AB = CD, we can use statement 4 to justify our writing AB = CD, On the other hand, if we know AB = CD and wish to establish AB s CD t we can use statement 3 for justification. The two statements, J and 2, or alternatively 3 and 4, are called converses of each other. That is, the statement "If p, then q" is called die converse of the statement "If q, then p" and "If q, then p" is the converse of "If p.. then q" The converse of a statement in the "if-then" form can be obtained by interchanging the if- and then-clauses. If p r then (j. If g, then p. For example, the converse of the statement: "If 1 live in Seattle, then I live in the state of Washington" is the statement "If I live in the state of Washington, then I live in Seattle." 198 Congruence of Triangles Chapter 5 It is evident from this example that a statement and its converse need not both be true. When a definition is given in the "if-then" form, it is understood mat the statements of the definition and its converse are both true. As an example, consider Definition l.l of Chapter 1: Space is the set of all points. This definition can be restated in the "if-then" form as follows: 1. If S is space, then S is the set of all points. The converse of (1) is 2, If S is the set of all points, then S is space. Thus the complete definition is: If S is space, then S is the set of all points; and if S is the set of all points, then S is space. The "if and only if' form (which is logically equivalent to the complete definition) is; S is space if and only if S is the set of all points. Although a definition in * "if-then" form always implies the converse statement, this is certainly not true of all postulates and theorems. We discuss the "if-then" form of a theorem more fully in Section 5.4. As an example of a postulate whose converse is not true, consider Postulate 2 (The Line-Point Postulate). "Every line is a set of points and contains at least two points." We can restate this postulate in the "if-then" form as follows: "If I is a line, then / is a set of points and contains at least two points." The converse of this statement is "If I is a set of points and contains at least two points, then t is a line." Clearly, this last statement is false. There are many sets of points such as planes that contain at least two points and that arc not lines. Consider the following theorem proved in Chapter 4. "Vertical angles are congruent." In the "if-then" form, this theorem can be stated as follows: "If LA and L B are vertical angles, then LA === LB" Surely the converse of this theorem, staled as follows, is false: "If LA=* LB, then LA and / B are vertical angles." 5.3 "If-Ttwn" Statements 199 We conclude this section with some remarks about definitions and truth. A definition in our formal geometry is accepted as a true state- ment. Why is "Space is the set of all points" a true statement? It is true "by definition." Definitions help us communicate. It is helpful to have one word that means the same thing as "the set of all points/* It is help- ful to have one word to describe several points that all He on the same line. Why do eollinear points all lie on the same line? They do, by definition. EXERCISES 5.3 In Fxerdses 1-5, a definition iu "if-then" form is given. Write its converse. L If the points of a set are eollinear. then there is a line which contains all of them. 2. If there is a plane which contains all the points of a set, then those points are coplanar, 3. If point A is between points B and C, then rays A3 and A C are opposite rays. 4. If an angle is a right angle, then its measure is 90. 5. If W is the midray of LABC. then BP is between BA and §C and ZAflFSB IFBC ' In Exercises 6-8, write the definition of the given phrase using the "if and only if form. 6. Acute angle 7. Linear pair 8. Vertical angles In Exercises 9-16, is the statement true or false? Write the converse of the statement. Is the converse true or false? 9. If two sets are convex, then their intersection is convex. 10. If two angles are right angles, then they are congruent. 11. If two angles are complements of congruent angles, then they are congruent, 12. If 5» is the interior of an angle,, then S is a convex set. 13. If two angles are congruent, then they are supplements of congruent angles. 14. If I! st CD, then WstAB. 15. If AABC^ AKLM, then AKLM St A ABC, lft If AABCss ADEF and AD£Fs ARST, then A ABC as ARST. 200 Congruence of Triangtos Chapter 5 5.4 THE USE OF CONDITIONAL STATEMENTS IN PROOFS As you have seen, many of the theorems have the form "tip, then </," where p and q are statements. Not all theorems are stated in this way, however, because it is sometimes easier to state them otherwise, but every theorem can be rostated in the "If p, then q" form. A statement of the form "If p> then tf is called a conditional The if-clause (the p statement) Is called the hypothesis and the then-clause (the q statement) is called the conclusion. In order to understand math- ematical proof better, we examine how such statements are used in proofs- Let us consider the following statements concerning two numbers x and y about which we know nothing except what we are told in the statement (A). (A) lix= y, then x 2 = y*. We see that (A) involves two statements. (B) x = y (the hypothesis) (C) x 2 = y 2 (the conclusion) Even though we do not know the replacements for x and y, can we say anything about the truth of statements (A), (B) s and (C)? Do we know that ( B) is true? Do we know that (G) is true? What about statement (A)? Your experience in working with numbers should help you to see that even though (B) need not be true and that (C) neeA not be true, (A) is true. Thus a conditional may be true even though its hypothesis and conclusion are not. Replace x and y with several pairs of numbers, some of which are equal and some of which are not. You should find that in those cases where you chose unequal numbers for x and y t both (B) and (C) are false and in 'those cases where you chose equal numbers for x and y, both (B) and (C) are true. But, in every case, (A) is true. Let us examine one such case where the replacements for x and y are unequal numbers. Suppose we replace x with 2 and y with 1 , Then statement (B) becomes 2 = 1 and statement (G) becomes 2 2 = I 2 , or 4 = 1. Of course, both statements (B) and (C) are false. However, if we accept the hypothesis that 2 = 1, use the multiplication property of equality to write 3 • 2 = 3 • 1, or 6 = 3, and use the subtraction property of equality to write 6 -2 = 3- 2, or 4= 1, we have shown that the statement "If 2=1, then 2 2 = I 2 " is true. Of course, a general statement in the form of a conditional is of no value in a specific situation if the hypothesis of the conditional is false 5.4 Conditional Statements in Proofs 201 in that situation. The truth of "If p, then (f does not by itself guarantee tlie truth of either p or q. But the truth of the conditional and of the hypothesis is a different story, as we shaU see. Let us go back, then, to the three statements (A), (B), and (C) in- volving x and y. Suppose that (A) and (B) are both true. ITiat is, suppose that the conditional and the hypothesis are both true. Then it follows logically that the conclusion (C) is true. Check this with our example. This is a most important concept in mathematical proofs. It means that we can assert (C) after we have proved or know that both (A) and (B) have been established On the other hand, it does not mean that (B) follows from (A) and (C). (In our example, x = y does not follow from x 2 = ry 2 , since we could also have x = — y<) In general,, if a conditional and its hypothesis are both known to be true, then the conclusion of the conditional is also true. More concisely, if we know that the following two statements have been established; L if p. then q then we may conclude that q has been established. This means, in our example, that if we know the following two statements are both true: (A) if as = y, then x 2 = y 2 (B) x=y then we can conclude x 2 = y 2 . How are you to know when a conditional is true? In our example, the conditional if x = y, then x 2 = y 2 is a theorem that can be proved using the properties of numbers. In our geometry, a conditional is accepted as being true if it is a postulate, a previously proved theorem, or part of a definition. Let us look at two more examples. Example I We know that the conditional if JB m CD, then AB = CD is true. Why? Suppose that we also know or have been able to establish that AB ^ CD. What can we conclude? 202 Congruence of Triangles Chapter 5 Fjxamph 2 We know that the conditional if two angles are right angles, then they are congruent is true. Why? Suppose that you are able to establish that Z A and LB are right angles. What can you conclude? If, in connection with the same conditional, you are able to establish that LA^ LB, can you then conclude that £ A and Z B are right angles? Why? EXERCISES 5.4 Write the theorems in Exercises 1-6 in "if-then" form. State the hypothesis and the conclusion of each. 1, Supplements of congruent angles are congruent. 2, Right angles are congruent H. Vertical angles are congruent. 4. Two angles of a linear pair are supplementary. 5. The intersection of two convex sets of points is a convex set. 6. The interior of a triangle is a convex set. In Exercises 7-10, a statement p and a statement q are given. In each exer- cise, write the truth value {'that is, true or false) (a) for the statement "If p, then q" and (b) for the statement "If q t then p" The answers to Exercise 7 have been given as a sample, % p: ml A = 90 and mLB = 90 q; LA^LB (a) T (b) F 8. p 9. p 9 10. p VB is the midray of Z AVC. ZAVBsb LBVC A-M-B and AM -a MB^ M is the midpoint of 3B. Z A and Z B arc supplementary angles. (f LA and Z B are a linear pair of angles. In Exercises 11-2G, certain given statements are to be accepted as true. Then a conclusion is stated. In each exercise, state whether the conclusion is true, false, or inconclusive (that is, not enough information is given to decide whether the conclusion is true or false). 11, Gkmi: x + y = 16, x - y = 12. Concision: x = 15, 12, Given: If there is not u cloud in the sky, then it is not raining. There is not u cloud in the sky. Conclusion: It is not raining. 5.5 Proof* in Two-Column Form 203 13. Given: If iherc is not a cloud in the sky, then it is not raining. It is not raining. Conclusion: There is not a cloud in the sky. 14. Given; Yon urc a member of the team only if you obey the training rules- Ken is a member of the team. Conclusion: Ken obeys the training rules. 1 5. Given; You are a member of the teai 1 1 only if you obey the training rules. Bill obeys the training nilei. Conclusicm: Bill is a member of the team. 16. Given: 2% + 3y = 12, x - y = 4 Conclusion: 3x + 2u = 16 17. Given: The intersection of two convex sets of points is a convex set. S and T are convex sets of points. S f\ T = H. Conclusion; R is a convex set. 18. Given: If a and b are numbers and if ab = 0, then a = or b = 0. a and b arc numbers, ab = 0, and a =£ 0. Conclusion: h = 19. (.'teen: If mLA = 30 and mZB s 00, then LA and Z B are complementary angles. ZA and LB are complementary angles. Conclusion: mL A = 30 and fnZB= 60 20. Given: Linda will marry Joe only if he will buy her a new house. Joe will buy Linda a new house. Conclusion; Linda will marry Joe. 5.5 PROOFS IN TWO-COLUMN FORM Students often ask, "What is a correct proof?" Unfortunately, there is no simple answer to the question. Making correct proofs is something that each of us learns by experience. A proof that may seem convincing to you may not be at all convincing to another person with much less experience in geometry than you. A deductive proof of a theorem is a set of statements, one or several or many, that shows how the conclu- sion follows logically from the hypothesis. To make a good proof it is important to think clearly about what is given and what is to be proved, and to consider various possibilities of statements which will lead from what is given to what is to be proved. It will help you considerably if you have a firm understanding of the postulates, definitions, and theorems already stated or proved. The proof of a theorem is often given in paragraph form. We have used this form of proof for most of the theorems proved thus far. For illustrative purposes look at the proof for Theorem 4.13 in Chapter 4. 204 Congruence of Triangle* Chapter 5 THEOREM 4.13 Any two right angles are congruent Proof: Every right angle has a measure of 90, Hence all right angles have the same measure, and hence they are congruent to each other. This proof consists of two sentences. Since the second sentence consists of two parts, there arc really three steps in the proof. These three steps, or links, form a chain of reasoning that shows how the conclusion, they are congruent, follows logically from the hypothesis, two angles are right angles. In writing this proof we did not provide reasons to support these three steps because we felt that the proof as given could be understood by someone with your background in geometry. If we were trying to con- vince someone with less practice, it would be necessary to give addi- tional statements for justifying each of the statements in the proof. One of the advantages of the paragraph type of proof is that it is not always necessary to give the reasons for all the statements when those reasons are obvious. However, when you give a proof in this form, you should be prepared to fill in the reasons. A two-column proof of a theorem consists of a chain of statements written in one column with a supporting reason for each statement written in a second column. When a statement in this chain is estab- lished because it is part of the hypothesis of the theorem, we simply write "hypothesis" or * 'given" as the reason. Otherwise, a statement may have as its supporting reason a combination of a conditional and its hypothesis. As stated before, a conditional is acceptable if it fa a postulate, a part of a definition, or a previously proved theorem. The hypothesis of this conditional should have appeared as an earlier state- ment in tile proof. The conclusion of the conditional should apply to the statement that is being supported A two-column proof of Theorem 4.13 follows. Note that in reasons which are conditionals, we write the numbers of the statements in which we have established the hypothesis of the conditional. THEOREM 4.13 Any two right angles are congruent. RESTATEMENT; If Z A and L B are any two right angles, then Proof: Hypothesis; £ A and L B arc right angles. Conclusion: £A ~ LB 5.5 Proofs in Two-Column Form 205 1 . L A and Z B are right angles. 2 mLA = 90, mlB a 90 3. mZA = mZ£ 4. IA^ LB 1. Hypothesis 2. If an angle is a right angle (1), then its measure is 90, 3. Substitution properly of equality (2) 4. If two angles have equal meas- ures (3), then they are con- gruent We note several important points about this proof. 1. The proof is not complete until the last statement in the left- hand column is the same as the conclusion. 2. When a statement Is part of the hypothesis, we write "hypoth- esis" or "given" as its reason, 3. When a reason is in the "if-then" form, its hypothesis refers to an earlier statement or statements for support. For example, the if-clause of reason 2 refers to statement 1, However, the then- clause of reason 2 refers to statement 2. 4. When a reason is not in die "if-then" form, and it can he written in that form, then it must satisfy the requirements stated in (3). For example, reason 3 simply states: "Substitution property of equality (2)," We could also have stated reason 3 as follows: "If a, b t and c are numbers and if a = c and b = c (2), then a = h." 5. in proving the theorem we have not proved statement 4 con- sidered by itself as an isolated statement Rather we have proved the following conditional: If statement 2, then statement 4. Your teacher may permit you to list your reasons simply by identi- fying the postulate, definition, theorem, or property of equality which supports each statement. If this is the case, the proof of Theorem 4.13 might read as follows: Statement 1. L A and L B are right angles. 2. mlA = 90, m£B- 90 3. mLA = mlB 4. £A==* IB 1. Given 2. Statement (1) and the defini- tion of right angle 3. Statement (2) and the substi- tution property of equality 4. Statement (3) and the defini- tion of congruent angles 206 Congruence of Triangles Chapter 5 We have shown three examples of proofs of the same theorem. Which is the best proof ? The answer is that they are all good. The two- column proof reminds us that we audi be able to give a reason for ever)' statement we make, and it also makes it easier to see which hypothesis we accept to begin our proof. Which proof you choose may depend on whom you are trying to convince. Usually in writing a proof your ob- jective will be to convince your teacher that you understand the proof. Your teacher may want you to have experience in writing both the par- agraph type of proof and the two-column type. EXERCISES 5.5 1. In the following two-column proof several reasons are given in the "if-then" form. For each such reason indicate to which statement the if-clause refers and to which Statement the then-clause refers. THEOREM Supplements of congruent angles are congruent. restatement; If /. < r i s a supplement of Z a and if L. d is a sup- plement of Lb and if La == Lb, then Lc m L<L 4- Proof; Hypothesis; L c is a supplement of L a, L d is a supplement of Lb. Lam Lh Conclusion: Lc ^ Ld dXb 1. Z c is a supplement of L (t, Ld is a supplement of Z b, 2. mLc+ mLa = ISO mLd+mLb = 180 3. La== Lb 4. mLa = mLb 5. mLc + mLa = m/.d + fflZ b 6. mLc = mLd 7. Lczz Ld 1. Hypothesis 2. If two angles are supplementary, then the sum of their measures is 180. 3. Hypothesis 4. Tf two angles are congruent, then they have the same measure. 5. If x, y, and z are numbers and if x = y and z = y, then x = z. 6. If a, b, x, y are numbers and if a — b and x = «/, then x — a = y-b. 7. If two angles have the same meas- ure, then they are congruent. 5,5 Proofs in Two-Column Form 207 2. Identity each of reasons 2, 4, 5 r 6, and 7 in Exercise 1 as a postulate, definition, theorem, or property of equality. 3. Write a two-column proof for these theorems from Chapter 4, ;a) Complements of congruent angles are congruent. In Vertical angles are congruent. 4* Write a proof for the following theorem in (a) paragraph form and ib) two-column form. THEOREM If two angles are both congruent and supplemen- tary, then each is a right angle. 5, Write a two-column proof of the theorem that vertical angles arc con- gruent using (a) the definition of vertical angles, (b) the definition of a linear pair, (c) the theorem that if two angles form a linear pair, then they are supplementary, and (d) the theorem that supplements of con- gruent angles are congruent (Hint: In the figure, let La and Z c he a pair of vertical angles. Prove Z a s Zc>) In Exercises 6-11, you are to perform some experiments with physical tri- angles. You will need a ruler, a protractor, and a compass. You are tv use these experiments as a basis for formulating the next three postulates in Section 5.6. Do the required constructions and measurements carefully and answer all the questions before proceeding to the next section. 6. Construct a triangle, AAJ3C, in which AB = 2| in., AC = 1^ in,, and m/.A = 50, Measure the remaining three parts (BC, /_!&,/.€) of your constructed triangle and compare your measurements with those of two or three of your classmates. Are they nearly the same? 7. Construct a triangle, ARST, in which HS = 2 in., m Z R - 40, and mZ S = (it). Measure the remaining three parts { Z T, RT t ST) of your constructed triangle and compare your measurements with those of some of your classmates. Are they nearly the same? 8. Construct a triangle, APQR, in which EQ = 5 cm., FR = 4.5 cm., and RQ = 6 cm. (You may need to use a compass for this construc- tion.) Measure the three angles of your constructed triangle and compare your measurements with those of some of your classmates. Are they nearly the same? J). Construct a triangle, &DEF, in which mLD = 50, m£B = 60, and mZF = 70. Measure the three sides of your constructed triangle in centimeters. Compare your measurements with those of some of your classmates. Are they nearly the same? 208 Congruence of Triangles Chapter 5 10. Construct a triangle, &LMN, iii which m£L = 40, LM — 5 cm,, and M\ ! = 3.5 cm. (You may need to use your compass again for tills con- struction,) Measure Z M, L N, and LN of your constructed triangle and compare your measurements with those of some of your classmates. Are they nearly the same? 11. Using only your ruler, construct a triangle, AADE, which has no two of its sides congruent. Now construct AA'DE* (distinct from AADE) such that AADE a A'D'F!. (You need not restrict yourself to using only your ruler for this construction.) How many of the six parts of AADE did you use in obtaining AA'D'E'? Could you have obtained AA'D'E by using a set of parts different from those that you actually used? What is the least number of congruent parts of AADE needed to be sure that AA'D'E 1 is congruent to AADE? 5.6 THE CONGRUENCE POSTULATES FOR TRIANGLES We asked you to perform certain constructions in Exercises 5.5. We now wish to examine these con- structions in more detail, but first we need some definitions. In A ABC (Figure 5-4) we say that Z A is in- cluded by sides AC and "KB. Simi- larly, we say that side B€ is included by angles Z B and Z C Definition 5.2 An angle of a triangle is said to be included by two sides of that triangle if the angle contains those sides, A side of a triangle is said to be included by two angles of that triangle if the endpoints of the side are the vertices of those angles. _ln A ABC, shown in Figure 5-4, which angle is included by sides BC and BA? Which side is included by Z A and Z C? Were you able to answer these last two questions without looking at the picture of the triangle? Without looking at a pienre of ARST, state which angle is included by sides ST and RT, Which side is included by Z R and Z S? In Exercise 6 of Exercises 5.5, you should have concluded that all triangles having the given parts are congruent; similarly for Exercises 7 and 8. When tins is true, we say that the three given parts determine a triangle. In Exercise 6, the three given parts of the triangle were two sides and the included angle. In Exercise 7, the three given parts were two angles and the included side, and in Exercise 8 the three given parts were the three sides. In Exercises 9 and 10 ? however, you should have found that not all triangles having the three given parts are con- 5-6 Congruence Postulates for Triangles 209 gruent. How many triangles of different size can be constructed if only the measures of the three angles are given (assuming there is at least one triangle with angles having these measures)? How many triangles of different sizes can be constructed using the data of Exercise JO? How did you construct AA'HE' in Exercise 1 1? Make a list of the steps you used. Perhaps you have one of the following lists. list I 1. Draw AU^ 2. Draw LA' = LA, 3. DrawA'Fs 4. Complete the construction by connecting R' and j r 7withD ? F, List2 1. Draw ZA's LA 2, Draw A' if ss AD. 3, Draw LU zs, LD. 4. Complete the constmction by drawing the sides ot LA' and LD 1 long enough. List3 1. DrawATFss AD. 2. Draw an arc with A' as center and A'Ef as radius. 3. Draw an arc with 17 as center and D'E' as radius. 4. Complete the construction by connecting the intersection of the arcs to A* and U. A' D' a* n- A' d- For each list, the figure at the bottom shows what the construction looks like just before it is completed. Tjook at the first list. How many side measures are used? How many angle measures? Is the angle between the two sides? This com- bination oitwo sides and the included angle is abbreviated by the sym- bol S.A.S.; the correspondence ADE + — ► A'UE' is referred to as an S.A.S, congruence because we feel that this much information about congruent pairs is enough to guarantee that all matched pairs of parts are congruent. We make this conclusion formal in Postulate 23, Look at the second list. What combination of measures is used? This combination of two angles and the Included side is abbreviated A.S.A.; the correspondence ADE* — >A'D'E' is referred to as an A.S.A. congruence. We make this conclusion formal in Postulate 24, Look at the third list. This combination of three sides is abbreviated S.S.S., the correspondence ADE< — *A'D'E' is referred to as an S.S.S. congruence. We make this conclusion formal in Postulate 25. 210 Congruence of Triangtes Chapter 5 POSTULATE 23 (The S.A.S. Postulate) Let a one-to-one cor- respondence between the vertices of two triangles (not necessarily dis- tinct) be given. If two sides and the included angle of the first triangle are congruent, respectively, to the corresponding parts of the second triangle, then the correspondence is a congruence. (Sec Figure 5-5.) A ABC fl D ADEF hv the S.A.S. Postulate Figure 5-5 POSTULATE 24 (Tkts A.S.A. Postulate) Let a one-to-one cor- respondence between the vertices of two triangles (not necessarily dis- tinct) be given. If two angles and the included side of the first triangle are congruent, respectively, to the corresponding parts of the second triangle, then the correspondence is a congruence. (Sec Figure 5-6.) A ABC = ADEF by the A.S.A, Postulate Figure 5-6 POSTULATE 25 (The EMS Postulate) Let a one-to-one cor- respondence between the vertices of two triangles (not necessarily dis- tinct) be given. If the three sides of the first triangle are congruent, respectively, to the corresponding sides of the second triangle, then the correspondence is a congruence. (See Figure 5-7.) B D A ABC * ADEF by the &&& Populate Figure 5-7 5,6 Congruence Postulate* for Triingles 211 Note that there is no A.A-A. Postulate (Figure 5-8) and no S.S.A. Postulate (Figure 5-9). Also note how the results of Exercises 9 and 10 of Exercises 5.5 arc related to this statement. B D AABCgt ADEF AAHC £ ADEF Figure 5-9 Our experience with physical triangles suggests that it would be proper to include an S. A. A. Postulate (Figure 5-10). Actually, we do not need such a postulate. The statement that you might expect as a postu- late is, in fact., given as a theorem in a later chapter. We defer the proof because we need not only the congruence postulates but also a postu- late about parallel lines, which appears later, before we can prove the S.A.A. statement. Since it is easy to prove later, we have decided not to adopt it formally as a postulate. AAiJC := AJ>EF b> S.A.A. Figure 5-10 The A.S.A. and S.S.S, Postulates can be proved as theorems once the S.A.S. Postulate is assumed. The proofs are difficult, however, and so we have adopted these statements as postulates in order to make simpler the development of our geometry. 212 Congruence of Triangles Chapter 5 EXERCISES 5.6 In Exercises 1-16, like markings on the triangles indicate congruent parts. In each exercise, determine if a pair of triangles can be proved congruent. If a congruence can be proved, write a triangle congruence (in the form ABAC ss &FDE) and the postulate (S.A.S., A.S.A., S.S.S.) you would use to prove it. Exercise 1 has been worked as a sample. J. ABAC at A FDE by&A£. h D 3. i *i» 6. R 7. 5.6 Congruence Postulate! for Triangles 213 13, V 214 Congruence of Triangles 17. For A ABC and ADEF, Chapter 5 ■\B - EF, I As:- IE, AC^ED. Write a congruence between the two triangles. What postulate are you using? IS. For ARSUmd &GKL, Which two sides must be proved congruent if AJRl/5 == &LKG by the A.S-A, Postulate? 19. For AFQR aud AQSR, FRsiSn and PQ^SQ, Arc the triangles necessarily congruent? Why? Draw a figure, 20. For A ABC and A BCD, IABC=* IDBC and LACB== LDCB. Are the triangles necessarily congruent? Why? Draw a figure. In Exercises 21-30, two triangles appear to be congruent in the given figure. Iu cadi exercise, certain information is given about the figure. Assume that all points are coplanar and have the relative positions shown. (a) Copy and mark each figure, as was done in Exercises 1 - 1 (i, to show the given information. (b) If the given information is sufficient to prove the triangles con- gruent, state a congruence between the triangles and the postulate you would use (S. A.S., A.S.A., S.S.S.) to prove it. If the given infor- mation is insufficient, write "Insufficient." £1. M is thcmidpoinl of FQ. 22. A~D - EC at D. PR==QR D is the midpoint of EC. 5.6 Congruence Postulate? for Triangles 215 23. VA is the midray of IBVC. m£VAC=mlVAB 27. L-F-W N-E-M D is the midpoint of LM, LF = ME DF = DE N 24. Consider onlv APQR and ASRQ. P0 _L UK, SR 1 OT, and 25. Consider onlv ARTS and AR VS. KT^RV ST=sW m, STV = ml$VT mlRTV = mARVT L D M In Exercise 27, if, in addition to the information given for the figure, we also haveLA T = MN, is AFNDatAEXD? Why? (Hint: Use ND in your proof. This segment exists even though it is not shown in the figure.) N-H-M U is the midpoint of GK. IsG LGK MR 1 GK N 26. A-E-D, D-F-C B is the midpoint of AC. BE 1 AD RF 1 £ft AE = CF 30, TR ± VR ^ SR A-R-B 216 Congruence of Triangles Chapter 5 5 J USING THE S,A.S. r A.S.A., AND S.S.S. POSTULATES IN WRITING PROOFS In this section and lire following one, you will be asked to write your own proofs. In writing these proofs you will usually need to prove one or more pairs of triangles congruent by using the S.A.S., A,S,A.„ and S.S.S. Postulates. Therefore, in planning your proof, you should look for the opportunity to apply one of these postulates to some pair of triangles. We illustrate with some examples. Example 1 If M is the midpoint of AB and CD, then AC ^ ED. In starling to construct a proof for this statement, we first draw a figure which seems to fit the hypothesis. c Figure Ml q Figure 5-11 shows M to he the midpoint of segments A~R and CD, (Can you draw a different figure which shows the same information?) We have marked AM and JIB with the same number of marks since, by the definition of a midpoint, we know thai A^f s MB. CM and MD have been marked alike for the same reason. Before attempting to construct a proof of a theorem, it is helpful to have a definite plan in mind. In this example, we want to prove that two segments are congruent. We know that two segments are congru- ent if they are corresponding sides of congruent triangles. Our plan is to prove A AMC = ABMD in Figure 5-11. For completeness the fig- ure is included as a part of the proof, as it should be. Pmof: Hypothesis: M is the midpoint of AB and CD. Conclusion: AC = H75 (Plan; Prove AAMC £s &BMD.) 5.7 Using the S.A.S., A.S.A., and S.S.S. Postulates 217 Reason 1. M is the midpoint of AB and CD. 2.ZMe*m. CM a W) 3. £AMC=z I BUD 4 AAMCj^ ABAID 5. AC ^ tf D L Hypothesis 2. Definition of midpoint (1) 3. Vertical angles arc congruent 4. S.A.S. Postulate (2. 3) 5. If two triangles arc congruent (4), then their corresponding parts are congruent Note that there are essentially the following five steps to writing a geometric proof in two-column form. L Draw a figure which seems to fit the hypothesis and, where pos- sible, mark on the figure the information given in the hypothesis. 2. State what is given (the hypothesis) expressed in terms of the figure. 3* State what is to t>c proved (the conclusion) also expressed in terms of the figure, 4, State a plan for the proof. The plan need not he expressed in written form, but should be carefully thought through before at- tempting to write the proof. 5. Write out the proof in two-column form, Now let us examine the proof for Example 1. You probably noticed in the statement of Step 3 that we relied strongly on the figure for giv- ing us the information that L AMC and Z BMD are a pair of vertical angles. If our figures are carefully drawn, we can rely on them to give us correct information, However, we must always be prepared to he able to justify any information that we take from a figure by postulates, definitions, and theorems. In the figure for the example, we know that points A, Af, B are eollincar and that points C, M, D are collmear (Why?). We also know that when two lines intersect, vertical angles are formed. Since the figure clearly shows all of this information, we did not bother to establish in our proof that I. AMC and Z BMD arc vertical angles. Since this may be your first experience in writing geometric proofs in two-column form, your teacher may want you to write a more com- plete proof than the one given. In other words, your teacher may not want you to assume any information from a figure without establishing this information in your proof. If this is the case, we give a second, more complete proof of the theorem in Example 1 for your consideration. 2 IS Congruence of Triangles Ifajjxithesis: M is the midpoint of AH and CD. Conclusion: AC =k BD Chapter5 1, M is the midpoint of A73 and CD. 2, M is between A and B, and M is between C and D. 3, Ml, M& and m£> m£> are two pairs of opposite rays. 4, IAMC and ZIttfD are ver- tical angles. 5, LAMCm IBMD e.A~M==MB,CM=*MD 7. AAMCss AJ3MD 8. A~€=±BD 1. Hypothesis 2. Definition of midpoint (1) 3. Definition of opposite rays (2) '1 Definition of vertical angles (3) 5. If two angles are vertical (4), then they arc eongnient. 6. Definition of midpoint (1) 7. S.A.S. Postulate (5, 6) 8. If two triangles are congruent (7), then their corresponding parts arc congruent, Note that some of the reasons are written in abbreviated form. For example, reason 4 is ''Definition of vertical angles'* rather than the com- plete statement of this definition. Be quite certain that you know the complete statement before using the abbreviated form in your proofs. Example 2 % in quadrilateral A#CD, TLB m BC and AB a CD, then la = ia Hypothesis: AD a 1$C 31 as ED Conclusion: LA^ LC (Plan: We want to prove L A = L C by showing they are corre- l ■ nding angles of congruent triangles. But there are no triangles in our figure. We therefore draw DB to show which triangles wc shall use/) S.7 Using the S.A.S.. ASA, and S.S.S. Postulates 219 Note that the segment JDH in Figure 5-12 is dashed to distinguish it from the parts of the figure given in the hypothesis. Figure 5-1S We call a segment such as DB an auxiliary segment. Thus an aux- iliary segment is a segment that is not a part of the figure given in the hypothesis, but does exist by the definition of a segment and the Point- Line Postulate. Such segments should be drawn into your figure when it is convenient to use them in the proof. We now continue with our plan. Since two sides of AABD are congruent to the corresponding sides of ACDB, we would expect to prove these triangles congruent by ei- ther the S.A.S, Postulate or the S.S.S, Postulate. If we were to use S.A.S., we would need ZA^ L C. But since this is what we are trying to prove, we cannot use it as part of our proof- That leaves the S.S.S. Postulate. In AABD, the third side is BD and in A CDB t the third side is DB. But BZ5 and DB are the same segment and are congruent by the reflexive property of congruence. We can now write the proof. 1. 35 s CD 2. AD ^ CB 3. Segment BD exists. 4 ED~.T5B 5. AABD =* A CDS a lAs IC 1. Hypothesis 2. Hypothesis 3. Point-Line Postdate and the definition of segment 4. The reflexive property of con- gruence for segments. 5. $.$£, Postulate (I. 2, 4) 6. If two triangles are congruent (5), then their corresponding parts are congruent. Note that when we wrote AD ^CSin statement 2 and BD s DB in statement 4, we were adhering to the correspondence ABD < — * CDB for the two triangles. Of course, it would be correct to write 315 m EC In statement 2 and BD m BD in statement 4. But it helps to keep tilings clear for both the writer of the proof and someone read- ing it if it is written the way we wrote it, keeping the order of the letters consistent with their order in the congruence we wish to prove. 220 Congruence of Triangles Chapter 5 Example 3 If points are as shown in Figure 5-13 and if AB & CB f FE ?~ DF, and ~BE ^ ED. prove AF = FC. How do we formulate a plan for this proof? A plan for a proof can often be formed by "working back- wards." That is, we begin with the conclusion and Lry to work our way back to the hypothesis, If these steps can then be reversed, we have a plan for our proof. This method LonsisEs ui writing ^or cnizuung information in two columns: or; A c Figure 3-13 I Can Prove If 1 Can Prove 1. JTszFC L (a) AE^CD? (b) FEzzUF\/ 2. JEseCO 2, AAB£ss ACBD? a AABE=*ACBD 3, (a) AB s CB (S))/ (b) ZB^Z/?(A)v/ (c) £EsAD(S)V read line 1 as follows: I can prove AF ^ FC if I can prove (a) AE at UD and (b) FE ^ DF. (Note that if we know statements (a) and iTj) of line 1 in the second column, then we ran deduce statement. I in the first column from Corollary 3.4.3.) Because (b) is given in the hypothesis, it is checked. Statement (a) is then to be considered. It is brought down to line 2. We read line 2 as follows: "1 can prove AE = <OD if 1 can prove A ABC ss ACBD" Since this triangle congruence is not given, we bring it down to line 3. Line 3 is read as follows: l T can prove A ABE ^ ACBD if I can prove (a) AB 3 CB, (b) IB == LB, and (c) HK "^ ED." Since all three of these statements are given or can easily be proved, they are checked. We can now write the proof by re- le order of the statements. Proof: Hypothesis: AH ^ CB FE^UF BE^ED Conclusion: AF ss FC Statement 1. BEsBD 2, ZBs IB 4. AABE = ACUD 5.AEaCD 6. FE a DF 7, AFaJU 5.7 UsingtKe S.A.S., ASA, and S.S.S. Postulate* 221 Reason 1. Hypothesis 2. Reflexive property of congru- ence for angles 3. Hypothesis 4. S.A.S. Postulate (1, 2, 3) 5. If two triangles arc congruent (4), then their corresponding parts are congruent, 6. Hypothesis 7. Corollaiy 3,43 (5, 6) Example 4 In Figure 5-14, all points are coplanar. Hypothesis: C is in the interior of L ADE. E is in the interior of L CDB. W IDE W is CD Copy and complete the proof that AE=zC!B. {Plan: We plan to use ADE /.ADEel LCDB by the Angle Measure Addition Postulate, and DE 3: UB. We can then use SAS.) Figure 5-14 -* CDB. We can show ID == CD, L KD _L DC, BD X DE 1. Hypothesis 2. ml ADC = 90, mlBDE = 90 2. [?] (1) 3. C is in the interior of Z ADE. 3. [?] E is in the interior of L CDB. 4. ml ADE = 90 + mlCDE mlCDB = 9() + mlCDE 5. mZAD£ = mZCDB 6. ZADEj^ZCDB 7. AD s= CD, DE s Dft 8. AAEEtt A CDB 9. 32 fi CB 4, Angle Measure Addition Pos- tulate (2, 3) 5, Substitution equality (4) & mo) 7* 8. {6, 7) 9. eke), m property of 222 Congruence of Triangles Chapter 5 EXERCISES 5.7 In Exercises 1-22, write two-column proofs. In each exercise, copy the fig- ure and mark on it the given congruences. L In the figure, S is between P and Q, BS ^ PQ, and Zla L 2. Prove that S is the midpoint of FQ. Copy the following outline and supply the missing reasons, includ- ing the numbers of supporting statements. Statement Reason 1. ll^JJl 2, BS=RS a RS ± PQ 4. jLHSPandARSQ are right angles. 5. /.RSF = £RSQ a APSil^ &QSB 7. PS == P 8. S is between F and Q. 9. S is the midpoint of P%). 2. In the figure, LARD and Z CBE are vertic al an gles, £ is the mid- point of DE, and ID St LE. Prove that B is the midpoint of 3. In the figure, point B is between points A and C and point C is between points B and D. Given CJR ^ DS, Zls Z 2, and AB m CD, copy and complete the proof that Zr =* IS. 1. Hypothesis 3. S 4. If lines determined by two seg- ments arc perpendicular (pi), then [7]. 5. Any two right angles are [?] 7. If {[UK then [g. am 9. HKSE) 5.7 Usng the S.A.S., A.S.A.. and S.S.S. Postulate* 223 1. B is between A and C, and C is between B and D, 2. KB s CD a EC*BC 7. AACrt a ABDS 1. Hypothesis ^ m a ru 4. Length-Addition Theorem for Segments (1, 2, 3) ''• 'ill 6. Hypothesis 8, OiiTCsponding parts of congru- ent triangles are congruent (0D- 4. In the figure, A is between C and D, B is be- tween C and K, K is (he midpoint of AB. L\ = Z 2, and AC = BC. Copy and com- plete the proof that LACK =z ABCK, L Zl a Z2 2. D-A-C and E-B-C 3. /, 1 and Z G&K form a lin- ear pair, and Z2 and Z CBK form a linear pair. 4. Z CAIC is a supplement of Z I, and Z CBK is a supple- ment of Z2. 5. ZCAKc-i ZCBK 0. K is the midpoint of XB. 7. [?]*»[?] 9. ttlsiBC 10. ACAK a AGBK 11. ZACK a ZfiCK 1.0 2, Hypothesis 3, Definition of linear pair (2) 4. If two angles form a linear pair (3), then they are supplemen- tary. 5, The supplements of congruent angles are \J]ih 4)« 6. rj] 7. Definition of midpoint (6) 8. Hypothesis 9. Definition of congruent seg- ments (8) io, rr|{[?],[?j,rT|) ILEKB) Chapter 5 224 Congruence of Triangtes 5. In the figure, LR = RN and LM = M-W Prove that mZL== m£N. A. Hypothesis: For A ABC, 5? is the midray of L ABC intersecting AT: in D. AC is the midray of I BAC intersecting EC in E. Points R, A, R r S are collinear in tliat order. L HAD m ZSB& Conclusion: A~E=*W) (Flan: Show that IS ss IS, that ZABE== /BAD (supplements of congruent angles); and that vilBAE = mlABD (halves of equals), Then use the AAA. Postulate to prove AABE B A BAD.) 7. GStant DA = CB CBXA^ To Prove: /L&m£C 8. Docs your proof for Exercise 7 de- pend on points A, B, C, and D being coplanar? ft Given: £lm Z2 Z3 = Z4 To Proce; PS = RQ and mZS= mZQ p 10. c liven: D and K are between A and B as shown hi the figure, zage« zjbcd dc=?EC lb Prone: AAC& =g A BCD AEsSD 1 1. Use the same hypothesis and figure given in Exercise 10 and prove (a) AACD m ABCE. (b) AS a TIE. 12, In the figure, E is in the interior of Z BA C, A-D-E, Z EDO' cs Z EDB, and BD s CU Prove that A~S is Lhemidrayof ZRAC 5.7 Using tht S.A.S., A.S.A,, and S.S.S, Postulates 13. Given; us AB = CB AD = CD To Prove: l\= £2 [Hint: DrawBD.) 14. In the figure, R, S, 7', and N are collinear in that order. £BSV=± ZFJTvMSs W t TTa VS. From L?=LV s N 15. In the figure, A-E-C, D-E-B, ZI» Z2, Z3ss Z4,and Prove that £C and £15 bisect each other at E. 16. In the figure, E is the midpoint of J5F, H is in the interior of Z DEC, G is in the interior of Z FEW. Z DEG s Z FEW, and TJEssGE. Prove that Z tf =* Z G- 17* Given the situation of Exercise 16, prove in two different ways that LDEIls* LFEG. 18. Given: Zl=* Z2 Af is the midpoint of R-S rt^st Prove: Z3=*Z4 £26 Congruence of Triangles Chapter 5 19. In thefigure^AD = BC, E is the midpoint of CD. and AE s BE, Prove that AC m& BD. A B Copy and complete the statements in the fallowing plan. I Can Prove If I Can Prove 1, AC s£ BD 2. A ADC's A BCD 1. AADCssS A BCD? 2. (a) [U-[2]{S)v' (b) ^=m{A)? (c) DC=CD{S)>J 3. AEDAs A£CB? 4 (a) ££^[7] ( C ) [U-[i](S)v \ T ovv reverse the steps in the second column and complete the proof. a LADC s /. BCD 4. A[?J & A[7] 20. In quadrilateral A BCD, M is the midpoint of AC and Z 1 Si Z2, Prove that any segment which contains M and has its endpoints in the inte- riors of CD and AB is bisected at M, [Hint: Draw EF such that E-M-F, A-E-B, and D-F-C. Now prove that AEMAmt AFMC and hence that EM = FA-/ . i 21. CHAUJ^CE PROBLEM. 1 1 1 A ABC T AC SS ATI Prove that the correspondence CAB *. — * BACisa congruence using the S.A.S. Postulate. Hence conclude that A CAB ^-- ABAC Does tills prove that IBsz Z C? Why? Does this prove that if a triangle has two congruent sides, then the angles opposite those sides are congruent? 5.8 Isosceles Triangle* 227 22. CHALLEN'CF PROBLEM. Ill AH ST, ZSsZF. Prove thul I he correspondence 5Tft « — * TSft is a congruence using the A.S.A. Postulate. Hence conclude _that ASTR m ATSfl, Does this prove that R$ =s JIT? Why? Does ihis prove that if a triangle has two congruent angles, then the sides opposite those angles are congruent? 5.8 ISOSCELES TRIANGLES Each triangle shown in Figure 5-15 has at least two congruent sides. Such triangles are called isosceles (from the Greek words iso and skelm meaning "equivalent" and " legs," respectively'!. Definition 5.3 An isosceles triangle is a triangle with (at least) two congruent sides. If two sides are congruent, then the remaining side is called the base. The angle opposite the base is called the vertex angle. The two angles that are op- posite the congruent sides are called the base angles. In A ABC of Figure 5-15, A B and ~KC are congruent sides. The base is JJQ the vertex angle is Z A. and the base angles are Z B and Z C. Name the base, vertex angle t and base angles of ADEF in Figure 5-15, If a triangle has three congruent sides as does &PQR in Figure 5-15, then any side may be considered as a base of the t dangle. The angle opposite a base is considered the vertex angle corresponding to that base, and the angles that include a base are called the base angles corresponding to that base. 228 Congruence of Triangles Chapw 5 Definition 5.4 A triangle with three congruent sides is called an equilateral triangle. A triangle with three congruent angles is called an equiangular triangle. If you worked Exercises 21 and 22 of Exercises 5.7, you already proved the following two theorems. THEOREM 5.1 (The Isosceles Triangle Theorem) The base an- gles of an isosceles triangle are congruent restatement: In A ABC (Figure 5-16), AC Si A75. Prove that ZDs LC. Flgur* 5-16 U (Plan: We will show that, for AABC, the correspondence BAC * — ► CAB is a congruence using S.A.S., and hence LB = LC by the definition of congruent triangles,) Proof: Statement 1. XCzzAB 2. ZAa LA 3. tt^XC 4. &BACm ACAB 5. LB^ LC 1. Hypothesis 2. Reflexive property of congru- ence for angles 3. Symmetric property of con- gruence for segments (1) 4. S.A,S. Postulate (1.2, 3) 5. Corresponding parts of con- gruent triangles (4) are con- gruent. Theorem 5.1 implies the result which follows. Its proof is left as an exercise. 5.8 Isoitttas Triangles 229 COROLLARY 5.1J If a triangle is equilateral, then it is equiangular. We know from Theorem 5.1 that if a triangle has a pair of congru- ent sides, then the angles opposite these sides are congruent. The eon- verse of Theorem 5.1 is also true, and we state it as our next theorem. THEOREM 5.2 (Converse of the Isosceles Triangle Theorem) If a triangle has two congruent angles, then the sides opposite these angles are congruent and the triangle is isosceles. bestatkmevt: In A DEF (Figure 5-17), LE^ LF, Prove that DF = DE and hence that A DEF is isosceles, F Figures 5-17 (Plan: We will show that, for A DEF, the correspondence EFD < — ► FED is a congruence using A.S .A. and hence DF » DE by the definition of congruent triangles,) Fnwf: 1. ZEss IF 2. EF^TE 3. ZF» IE 4. AEFD BS AFED 5. DF=sDE 6. ADEF is isosceles, 1. Hypothesis 2. Reflexive property of congru- ence for segments 3. Symmetric property of con- gruence for angles (1) 4. A.S.A. Postulate {1, 2 f 3) 5* Corresponding parts of con- gruent triangles (4) are con- gruent. 6, Definition of isosceles triangle (5) The proof of the following corollary of Theorem 5.2 Is left as an exercise. COROLLARY 5.2.1 if a triangle is equiangular, then it is eqi 2 30 Congruence of Triangles Chapter 5 EXERCISES 5.H h Is every equilateral triangle isosceles? Is every isosceles triangle equilateral? 2. In the fipnjre. F is in th<-' interior of A ABC, AB =* AC and FB ss PC Prove, without using congruent triangles, that ZABPftZACP. 3. Write a two-column proof of Corollary 5.1,1. 4. Write a proof, in paragraph form, of Corollary' 5.2,1, 5. Prove Theorem 5.1 using the S.S.S, Postulate. 6. In the figure, A RST is an isosceles triangle with vertex angle at H. Give a proof different from that given for Theorem 5.1 that base angles of an isosceles triangle are congruent (Hint: Let M be the midpoint of ST and prove 7. Given: F-Q-S, F-R-T, and £IS3 Z2 l^ve: &QPR is isosceles, 8. Civen: A-B-D t A-C-E, AH =s AT?, and Pnwe: AflDCsACES 9, challenge problem. Did you use the Isosceles Triangle Theorem in your proof of Exercise 8? If not, write a different proof, using the Isos- celes Triangle Theorem* If you did, write a different proof in which the Isosceles Triangle Theorem is not used. 10. Given; A/)£F is isosceles with vertex angle at D, Prove: (ft) AGDH is isosceles, (b) Z UGH =s I DUG 5.8 lio&celes Triangle* 231 D K G H F 1 1. If points A, B, C, A F, F have the betweenness relations shown in the figure, if A ABC is isosceles with vertex angle at li, and if A AFC £s isosceles with vertex an- gle at F* prove AD as EC (Phn: Use the Isosceles Triangle The- orem to prove A A DC fe ACEA by A.S.A.) 12. In convex quadrilateral ABCD, JS^AT) and 55 a; W< (a) Prove I ABC ^ I ADC widiout proving any triangles congruent, (bj Draw AC intersecting B~D at E. Prove BE = Wand AC ± ED. 13w In the figure, Z 1 ^ Z 2, points B, I>, F„ C are collinear. Prove that AABC is isosceles. BD = EC 232 Congruence of Triangles 14. Given: ARST is equilateral. Prove: ARST is equiangular. (Plan: Show that the correspond- ence ilST « — » TRS is a congruence by S.S.S, and that IR=*IT&£S by the definition of congruent triangles.) t5. Given: APQR is equilateral, with A, R. and C the midpoints of i 3 ^, ^H, and M, respec- tively. Prove: AARC is equiangular. Chapter 5 1 0, challkn Gg l'noiiLEM. We can prove the A . S. A. Post ill ate as a theorem once we have assumed the S.A.S. Postulate. Complete the proof of the following statement. Given an A.S.A. correspondence FAM < — » RSP as indicated in the figures, if ZFs Zfl, IFAM== ZS, prove that A FAX! ^ A RSP. Proof: There is a point If on FM such that FM' ~~. HP. Why? (Note that our figure shows ,\f and Sf to be different points. We will prove- that they are the same point.) Therefore AFAM* ss. A RSP by S.A.S- {Show this.) IFAM' S I RSP. Why? Therefore AA$' = A*f by the Angle Construction Theorem, It follows that M f = M (Why?) and AFAM m ARSF. 5,8 lsosc«le* Triangtes 233 17, ciiam-Ekce problem. We can prove the S.S.S. Postulate as a theorem once we have assumed the 5.A.S. Postulate and proved the Isosceles Triangle Theorem. Complete the proof of the following statement. Given an SS.S. correspondence ABC <-h> PQH as indicated in the figure, if AC =r VH, prove that A ABC a APQR. 1. AB = FQ t B~€~i §R r Wmm 2. There is a point F on the opposite side of An from C such that ABAFr- £P> 3. There is a po int D on AF such thatAD^ PR. t &ABD- s\PQR 1. Given 2. Angle Construction Theorem 3. Segment Construction The- orem 4. 5.AS Postulate (1,2, 3) Since C and D are on opposite sides of AB, CD intersects AB in a point H. Our figure shows H to he between A and B. We could have, however, A = il.avB = H, or A -B U, or H-A-B, Complete the proof for the case where A-H-B by using the Isosceles Triangle Theorem and the S.A.S. Postulate to prove that A ABC st AABD, It will then follow from step 4 and the transitive property of congruence for triangles thai A ABC" = &.PQR* Draw a figure and complete the proof for the other cases, that is, for A = 1L B = H, A-B-H, and II-A-B, 234 Congruent* of Triangles Chapter 5 5.9 MEDIANS AND PERPENDICULAR BISECTORS In Figure 5-18, M is the midpoint of side W of A ABC Segment AM is called a median of A ABC. Since each side of a triangle has exactly one mid- point, every triangle has exactly three medians. Draw a triangle and its me- dians. What property do the medians appear to have? Figure .MS Definition 5.5 A median of a triangle is a segment whose endpoints arc a vertex of the triangle and the midpoint of the side opposite that vertex. THEOREM 5.3 The median to the base of an isosceles triangle bisects the vertex angle and is perpendicular to the base, restatement: In A ABC (Figure 5-l c r, AB == AC and AM is the median to BC. Prove: L AM is the midray of Z BAC. 2. AM 1 W. Proof: Copy and complete the following proof. 1. 7&mM 2. ZBs= LC 3. AM is the median to EC 4* M is the midpoint of EC, 6. AABM^AACM 7. ZBAAf == /.CAM 8. Af is in the interior of Z BAC, 9. AA? is the midray of L BAC. 10. /.BMA= I CM A 11. IBMA and Z.CMA are a linear pair. 12. Z BAM and Z CAM are right angles. 13. AM _L BC 1. Hypothesis 2. \t}{\2}) 3. Hypothesis 4.0(1 8. Theorem 4.11 (4) 9. Definition of midray (7, 8) io. 03 (00) 11. Definition of linear pair (4) 12. Theorem 4.12(10, 11) 13. 0(0) 5,9 Medians and Perptndicular Bisector* In Figure 5-20, A ABC is isosceles with vertex angle at A. Ray AG is the midray of the vertex angle. It appears from the figure that AG bisects BC and is perpendicular to EC at M. This brings us to our next theorem. 235 Figure 5-20 THEOREM 5.4 The midray of the vertex angle of an isosceles triangle bisects the base and is perpendicular to it. Proof: In proving this theorem it is necessary to show that ray AG intersects BS in a point M that is between B and C (as Figure 5-20 sug- gests), By the definition of midray, AG is between rays A/5 and AC. By the definition of betweenness for rays, B and C are in opposite half- planes with edge AG. By the definition of opposite sides of a line, there is a point Af of AG between B and C, Since M is between B and C, it follows from Theorem 4.11 that M is in the interior of L BAC. Since M is in the interior of L BAC and since it is a point of AG or of opp AG, but not both, it follows from Theorem 4. 1 that M is a point of AG. The rest of the proof of Theorem 5.4 is straightforward and is left as an exercise. In Theorem 5.4, it was required to prove that a certain ray was per- pendicular to a certain segment and that the ray bisected the segment The line that contains the ray and that is in the same plane as the seg- ment is called the perpendicular bisector of the segment in that plane. We state this formally in the following definition. Definition 5.6 The perpendicular bisector of a segment in a given plane is the line in that plane which is perpcndiculat to the segment at its midpoint. 236 Congruence of Triangles Chapter & In Figure 5-21, line I is perpendicular to A73 at M t the midpoint of A~B. Since I and AB determine exactly one plane (Why?), we say that / is the perpendicular bisector of AB in that plane. We sometimes write "I — bisA~B" for "/ is the perpendicular bisector of A~B." M Figure 5-21 Note that, in space, there is more than one line (How many?) that is a perpendicular bisector of a given segment. However, in a given plane the perpendicular bisector of a segment is unique, since a seg- ment has exactly one midpoint and, by Theorem 4.14, in a given plane there is exacdy one line that is perpendicular to a given line at a given point on the line. The next two theorems serve to characterize the set of points in the perpendicular bisector of a segment, THEOR&W 5.5 (The Perpendicular Bisector Theorem) If, in a lhvcu plant* a. Pis a nnitil on the perpendlCMbl I fceGtOE of AHk tfagB P is equidistant from the endpoints of AR restatement: In Figure 5-22, P is a point on line / in a plane a. I JL bisAB at M in a, l Prove: PA - FB Proof: If P = M, then PA = PB by the definition of midpoint. If P is any point in I different from M, then &APM £= ABFM by S.A.S. {show this); hence PA == ?B by the definition of congruent triangles and PA = PB by the definition of congruent segments. Figure 5-22 5.9 Medians and Perpendicular Bisectors 237 THEOREM 5,6 (Converse of the Perpendicular Bisector The- orem) If, in a gpven plane a , P is equidistant from the endpoints of AB t then P lies on the perpendicular bisector of XB, RESTATBMKNT! In plane a, PA s PB, M is the midpoint of AB. Prove: P is on I, the perpendicular bisector of AB. l Proof; If P is on line AB, dien P = M tacause AB has only one mid- point, In this case, P is on line I by the definition of the perpendicular bisector of a segmen t, I f P is not on line AB, then AAPM =s A BPM by S.S.S. (show this). Therefore £AMP is a right angle (Why?) and PM ± bis AU. Since, in a given plane, a segment has only one perpendicu- lar bisector, P is on I, EXERCISES 5.9 L Copy and complete the proof of Theorem 5.3. 2, Complete the proof of Theorem 5,4 by writing it in two-column form. You may assume that it has been proved thai AG intersects J5C in a point M which is between B and C, (See Figure 5-20.) 3. Copy A ABC below. (a) Construct the median from A to ISC. (b,i On the same figure construct the midray of L BAC. (c) Does the midray contain the median? {d) What must be true about A ABC if the midray from A is to coincide with the median from A? 238 Congruence of Triangles Chapter 5 4. Copy A DEF in the figure. (a) Construct the perpendicular bisector of EF in plane DEF', (b) Does the perpendicular bi- sector of EF contain point D? (c) What must be true about ADEF if _the perpendicular bisector of EF is to contain D? 5. From which theorem may we deduce that the vertex of the angle op- posite the base of an isosceles triangle lies on the perpendicular bisector of the base? 0* In the figure. A, B, C, D are distinct coplanar points, KB St /03, and (a) Is BD tile perpendicular bisector of AC ? (b) Is AC the perpendicular bisecLor of BD? (c) Which auxiliary segment is needed to prove AABCm AADCf (d) How do we know this segment exists? (e) Why is IB^ID? (f) If your answer to (a) is Yes, prove it is correct with a paragraph style proof. Do not use congruent triangles, (g) If your answer to (b) is Yes, prove it is correct with a paragraph style proof. Do not use congruent triangles. 7. In the figure, / is the perpendicular bisector of FQ. If the lengths of .seg- ments are as marked, find a, h, c, and d. 12/ / / V \J \ K 8 c \ rf%. y^ 5.9 Medians and Perpendicular Bisectors 239 I In the figure, A ABC is isosceles with AR = BC. If BD is a median, prove that AABD s ACBD. h In the figure, A PQR is isosceles with PQ b HI, The midrays of L QRP and L PQR intersect at T. Prove thai P? 1 QR. Do not use congruent triangles in your proof. 10. If t _L bis A~B at M, and if P is a point on / different from M, prove that FKI is a median of A PAR. 11. If S is the midpoint of (JR andPS 1 pR, prove that APQR is isosceles. (It. is not necessary to use congruent triangles in your proof,) 12. Given the figure as marked with F the midpoint of A~B, prove that DFl IE 13. < itALLtvGt problem. If AB and ($R are coplanar and congruent seg- ments, n 1 btsA~Q, m 1. bis BR, and mC n = P, prove that AABP =* AQRF. 240 Congruence of Triangles Chapter 5 In Exercises 14-22, you will need to prove certain triangle* congruent that arc not necessarily coplanar. You should look at the figures in perspective and be aware that angles or sides that are congruent may not look congruent in the figures. In every case, carefully draw a figure on your own paper and mark on it the congruent parts before writing a proof. 14. In the figure, points A, B, and C are in plane a. Point P is not in plane a. TB _ AB and FB 1 BC. If L BAC = £BCA t prove A~P ss UP, 15, In the figure, points D, E, and F are in plane /?. Point R is not in plane £ If RD jl DE f KD _ DF, and DE^W, prove A REF is isosceles and that LREF ^ £RFE. 16. In dihedral angle A -RS H. AUis in plane a, BC is in plane /?, ZC ± iT5 r HC i RS. D is in /IS. and \A('.B is isosceles with vertex stl C. Prove that A BDA is isosceles. 5.9 Medians and Perpendicular Bisectors 241 17. Gives TR, W, and KT bisect each other at S. Prove AKDP =± AT/R A ^— ^ 18. In dihedral angle E-AB-F, E is in plane ft, K is in plane /J, and AB con- tains C and D, If triangles DEC and DFX.' are isosceles with vertex an- gles at £ and F, respectively, prove that L EOF == Z ECF, (Plan: Prove AEDF s AECF by S.S.S.) {See figure above at right) 19. In Exorcise 18, is DC l>etween rays DF and DE? Can we prove Z.EDF& Z ECF by the Angle Measure Addition Postulate? 20. In the figure, PA = PB = PC and lAPB=zlBPC=z I A PC. Prove that AABCte equilateral. 21. In Kxercise 20, if we accept the hypothesis (but disregard the figure}, is it possible that P, A t B> C are coplanar? Try to draw a figure for the "plane case," 22. caiAjxENCE problem. In the fig]:. /' ;iru.l pare on opposite sides of plane a which contains points D, E, and EUPQ ±DK at E,TQ ± FE at E, and M == @E, prove &FDF ^ A QDF. 242 Congruence of Trfangles Chapter 5 CHAPTER SUMMARY The Definitions If t for A ABC and ADEF. L A *. — » ZD, KB < — » DE, IB < — * IE, W < — > EF, I C * — ► IF, AC < — > JJF y we indicate this correspondence by writing ABC <—— * DEF where the or- der in which the vertices arc named preserves the six correspondences. We speak of the pairs In the six correspondences named as CORRESPONDING PARTS of the two triangles. Two triangles (not necessarily distinct) arc CONGRUENT if and only if there exists a one-to-one correspondence be- tween their vertices in which the corresponding parts arc congruent Such a one-to-one correspondence between the vertices of two congruent trian- gles is called a CONGRUENCE. An ANGLE of a triangle is said to be INCLUDED by two sides of that triangle if the angle contains those sides. A SIDE of a triangle is said to be I N( .1 .UDED by two angles of that triangle if the cudpoints of the side are the vertices of those angles. The combination TWO SIDES AND THE INCLUDED ANGLE is abbreviated by the symbol S. A«S, If there exists a correspondence between two triangles such that S.A.S. of one triangle are congruent to S.A.S, of the second triangle, then we call this an S.A.S. CONGRUENCE, The combination TWO ANGLES AND THE INCLUDED SIDE is abbreviated by the symbol A.S.A. If there exists a correspondence between two triangles such that ASA. of one triangle are congruent to ASA, of the Second triangle, then we call this an ASA. CONGRUENCE. The combination THREE SIDES is abbreviated by the symbol S.S.S. If there exists a correspondence between two triangles such that S.S.S. of one triangle are congruent to S.S.S. of the second triangle, then we call this an S.S.S. CONGRUENCE. If a triangle has (at least) two congruent sides, it is called an ISOS- CELES triangle. The VERTEX ANCLE of an isosceles triangle is the angle included by the congruent sides. The BASE of an isosceles triangle is the side that is opposite the vertex angle. The BASE ANGLES of an isosceles triangle are the angles whose vertices are the endpolnts of the base. An EQUILATERAL triangle is a triangle that has three congruent sides. An EQUIANGULAR triangle is a triangle that has three congruent angles. A MEDIAN of a triangle is a segment that has for its endpoints a vertex of the triangle and the midpoint of the side opposite that vertex, In a given plane, the PERPENDICULAR BISECTOR OF A SEG- MENT is the line that is perpendicular to the segment at its midpoint. A statement of the form "If fh then </" is called a CONDITIONAL. The statement p is called the HYPOTHESIS of the conditional and the Chapter Summary 243 statement q is called the CONCLUSION. If a conditional and its hypothesis are known to be true, then it follows logically that its conclusion is also true. The statement **ff q, then p" is called the CONVERSE of the statement "If p, then q" and the statement "If p t then q" is called the converse of the statement "If q y then p" A definition in "if-then" form is to be understood as a conjunction of two statements "If p, then q" and "If q t then p" Some- times this is abbreviated to "p if and only iiq," The Postulates There were three postulates in this chapter having to do with congru- ent triangles. We list them In their abbreviated form only. Be sure that you know the complete statement of each postulate. POSTULATE 23. The S,A<S. Postulate. POSTULATE 24. The \SJi. Postulate, POSTULATE 25. Tlie S.S.S. Postulate. The Theorems There were six tlieorems and two corollaries in this chapter. We list them in the order in which they appeared. Re sure that you know what each theorem says and that you understand its proof. THEOREM 5 J (77ie hosceles Triangle Theorem) The base angles of an isosceles triangle are congruent. COR OLLAR Y 5.1.1 If a triangle is equilateral, then it is equiangular. THEOREM 5.2 (Converse of the I*o$eek* Triangle Theorem). If a triangle has two congruent angles, then the sides opposite these angles arc congruent and the triangle is isosceles, COROLLARY 5,2. 1 I! a Lrifuigle is equiangular, then it is equilateral THEOREM 5.3 The median to the base of an isosceles triangle bi- sects the vertex angle and is perpendicular to the base* THEOREM 5.4 The midray of the vertex angle of an isosceles tri- angle bisects the base and is perpendicular to it. THEOREM 5.5 (The Perpendicular Bizector Theorem) If, in a given plane a, P is a point on the perpendicular bisector of KB, then P is equidistant from the endpofnts of KB. THEOREM 5.6 (Converse of the Perpendicular Bisector Theorem) If, in a given plane «. P is equidistant from the endpoints of KB, then P H*s on die perpendicular bisector of KB. 244 Congruence of Triangles Chapter 5 REVIEW EXERCISES 1. Write the fallowing statement as a conjunction of two statements in the "if-then" form: A triangle is equilateral if and only if it is equiangular, 2. In your answer to Exercise 1, what is the second statement called with respect to the first? Are both statements true? 3. Write the converse of the statement: If two triangles are congruent, then their corresponding parts are congruent 4. In Exercise 3, are both the statement and its converse true? 5. Write the convene of the statement: Vertical angles are congruent. 6. In Exercise 5. is the given statement true? Is its converse true? ■ In Exercises 7 IS, decide if the sentence is true or false. 7. In the statement, "If p, then q" p is the conclusion and q is the hypothesis. 8. In the statement, "p only if q," p is the conclusion and q is the hypothesis, 9. In the statement, "p if tf," p is the conclusion and q is the hypothesis. 10. If we know the statement "If jt>, then q" is true and if we know that p is true, then we can conclude that q is true. 11. TJie statement "If p, then qr" may be true even though the statements p and q are both false, 12. If A is on the perpendicular bisector of CD, then AC = AD, 13. If, in a given plane, PE = PF, then P is on the perpendicular bisector of EF. 14. In a given plane, more than one line can be drawn which is both per- pendicular to a segment and bisects the segment. 15. In a given plane, more than one line can be drawn which bisects a segment. 16. If R, S t and T are not colLuiear and R is on the perpendicular bisector of ST, then A RST is isosceles, 17. Congruence of triangles has the transitive property. 18. In our formal development of geometry, we postulated that the sides Opposite a pair of congruent angles in a triangle are congruent. 19. Copy and complete: If, in a given plane, R is (?] from the endpoints of ST. then R lies on the (7] [3 of ST. 20. Copy and complete: If JAB is a segment and if R and S are disti n ct points such that R, S, A, H are coplanar, RA = RH, and M = SB, then jTj is the |7]|l] of A3- 21. If, la A ABC, ED bi sects Z ABC and intersects AC at D, ~KB m EC, prove that D in the midpoint of AC. 22, If, in the figure, A~B - BC, does it follow that lBADs*£BCD because base angles of an isosceles triangle are congruent? Why? 23, In the figure for Exercise 22, if ■\r. w. I BAD a . BCD. and D is in the interior of AABC t write a plan for proving A ABD ^ ACfJD. 24, Given: AC^BC lACD^ Z.BCE A-D-Ejy-E-B Prove; CD rr CK Revktw Exsrcises 245 B 25. Given: I'rnt r; A-D-C, B-E-Q B-P-D A-P-E, 11? EM RE LCDP& I CEP AAFB is isosceles. 26. Given: A3 = CD t A-M-Q B-XS-D.aBdADssEC Prove: (a) A ABD ss ACDB (b) ^lsZ2 (c) A A CD ss A GAB fd) Z3aZ4 (e) ACMD as AAJlfB (f) MJs the midpoint of BDandAT, 27. In the plane figure, \f is the midpoint of segments A~C and EP, BE sm Prove (a) AAMFss&CME. (b) AABC== ACDA. 28. Given: A-D-E, D-E-B AD"? BE CD s ce Prove: A ABC is isosceles. Prove, in order, Zl = /2 Z3= i. J AACDj^ ABCtf AC m BC A ABC is isosceles. Chapter Carl Stru wc/Monkmetjtfr Inequalities in Triangles 6.1 INTRODUCTION In Chapters 3 T 4, and 5 we developed the concept of congruence for segments, angles, and triangles. When we say that two segments are congruent, we mean that two numbers associated with these seg- ments, their lengths, are the same. The concept of congruence for an- gles and for triangles is based also on the concept of equal measures, and hence on the concept of equality for numbers. In this chapter we are concerned with segments and angles, in par- ticular with segments that are not congruent to each other and with angles that are not congruent to each other. In other words, wc are concerned with segments of unequal measures and with angles of un- equal measures. In comparing two segments {or angles) that are not congruent to each other, it is useful to know which one has the larger measure. To express such a comparison it is convenient to use some fa- miliar words and symbols defined formally as follows. Definition 6.1 XB > CD if and only if AB > CD< AB < CD if arid only if Afl < CD. Definition 6.2 I ABC > LDEFii and only if m I ABC > mLDEF, IABC< IDEF if and onlv 'if mlABC< mLDEE. 248 Inequalities in Triangles Chapter 6 When used to compare segments, the symbol "]> M may be read as "greater than," or "longer than," or "larger than." Similarly, **<" may he read as 'less than," or "shorter than," or "smaller than." When used to compare angles, the symbol "]>" may be read as "greater than" or "larger than." Similarly, "<*' may be read as "less than" or "smaller than," Note that Definitions 6, 1 and 6.2 express a comparison of segments, or of angles, in terms of an inequality involving numbers. It should be clear that an inequality involving segments or angles is really a state- ment of "not-congruence" together with a statement of which seg- ment, or angle, has the greater, or lesser, measure. Because the proper- ties of inequalities for numbers are essential for the development of in- equalities for segments and angles, we review them in Section 6,2. Tliis chapter includes several important theorems, some of which involve comparisons of parts of one Lriangle. Others involve compari- sons of parts of one triangle with parts of another triangle. To save time we shall, in some cases, give "abbreviated proofs" of these theorems, that is, just the key steps in the proofs. You should be able to supply a complete proof if asked to do so. 6.2 INEQUALITIES FOR NUMBERS We begin by defining* formally, less than and greater than for numbers. Definition 6.3 If a and b are numbers, then a K h if and only if there is a positive number p such that b — a + p. Also a > b if and only if there is a positive number p such that a = b + p. Definition 6.4 If a and b are numbers, then a < b if and only if a < b or a = b. Example 1 The statement 4 < 5 is read "4 is less than or equal to 5" and, by Definition 6.4, it means 4 < 5 or 4 = 5. We know that a dis- junction of two statements is true if either of the two statements is true. Therefore 4 < 5 is a true statement because 4 < 5 is true. Similarly, 5 < 5 (which means 5<5or5 = 5)isa true statement because 5 = 5 is true. Most of the numbers in our geometry are positive numbers and represent lengths of segments, measures of angles, measures of plane 6.2 InequalJttot for Numbers 249 regions (areas), or measures of regions in space (volumes). From Defi- nition 6.3 we can conclude that if x, y, and * are positive numbers and if x = if + z, then both tj and z are less than x. We get this by first con- sidering z as the positive numt>er p in the definition and then consider- ing y as the positive number p. This gives us Lhe two statements x = y + p and x — z 4- p« Thus, by Definition 6.3, y < x and z < x. That is, both numbers in a sum of two positive numbers are less than the sum. For example, 15 = 9 + a Hence 9 < 15 and 6 < 15. From the second part of Definition 6.3, we can conclude that the sum of two positive numbers is greater than either of them. For example, 32 = 21 -J- 11, Heneo32>2J and 32 > 11. The following theorem is easy to prove using Definition 6.3. THEOREM 6.1 If x and y are numbers, then x < y if and only if There are two parts to Theorem 6.1, 1 . If x and y are numbers, and if x < y, then y > x. 2. If x and y are numbers, and if y > x, then x < y. Proof of part b Statement Reason 1. x and y are numbers and 1. Hypothesis x<y. 2. y = x + p, where p is a posi- 2, Definition 6.3 tive number. 3. y > x 3. Statement 2 and Definition 6,3 JVoo/ of part 2: Assigned as an exercise. We now state five properties of order (inequalities) that are helpful in proving theorems about geometric inequalities. You may consider these properties as postulates for the real number system, although we could prove Properties 0-3, 0-4, and 0-5 by using Definition 6,3 and Properties Q-J and 0-2. 250 ln«qu*fties in Trianglss Chapter 6 0-1 The Positive Closure Property. If x and y are numbers, and x > and y > S then x + t/ > and xy > 0. 0-2 Trichotomy Properly. If x and t/ are numbers, then exactly one of the following is true: x < y> x = I/* x > j/, 0-3 Transitive Property. Tf a;, y, and z arc numliers, and if x <C J/ and |( < # then x < z. Also, if * > y and y > £, then % > % 0-4 Addition Property. If a, h, x, and y are numbers, and if x <C ^ and a < fe, then i + fl < y t &• Also, if x ]> t/ and a > i>, then x + a > y -+- fc. 0-5 Multiplication Property. If x, y, and z arc numbers and if x < y and s > 0, then xz <C ys. Also, if x >■ y .and z > 0, then We now prove several theorems which are useful in the sections that follow. THEOREM 6.2 AS > CD if and only if CD < "KB, There are two parts to Theorem 6,2. 1. If AB > CD, then £75 < "KB. 2. If CD < AB r then IB > CD. Proof of part 1: Statement Reason 1. £B>CD L Hypothesis 2. AB > CD 2. Definition 6.1 a CD<AB 3. Theorem 6.1 4. CD < ZB 4. Definition 6. 1 Proof of part 2: Assigned as an exercise, THEOREM 6.3 IABC> ADEF if and only If IDEF<: I ABC. Proof; Assigned as an exercise. THEOREM 6.4 Let three distinct collinear points A, B, C be given. Then A-C-B if and only if AB > AC and AB > BC 6.2 Inequalities for Kumbers 251 Tnere are two parts to Theorem 6.4. (Draw a figure showing the re- lationship between points A, B, and C.) J, Given three distinct collinear points A, B ? C, if point C is be- tween points A and B, then AB > AC and AB > BC. 2. Given three distinct collinear points A, B, C, if AB > AC and AB > BC^ then C is between A and B. Proof of part ! . Since C is between A and B, AB = AC -f BC, Why? But AC and BC are positive, Hence by Definition 6.3 AB > AC and AB > BC. % Proof of part 2; Points A. B, C are coUinear (given), so exactly one of them is between the other two. Why? That is, we must have exactly one of the following three belweenncss relations: B-A-C t A-B-C, or A-C-B. We shall show that the first two of these are impossible; hence the only remaining possibility is A-C-B. Suppose that A is l>etween B and C, Then, by the first part of Theorem 6,4, we must have BCy AB. But this contradicts the hypothesis that AB > BC; hence A is not be- tween B and C. Now suppose B is between A and C. Again, by part 1 of Theorem 9.4, we must have AC > AB and this contradicts the hy- pothesis that AB > AC; hence B is not between A and C. The only remaining possibility is that C is between A and B. This completes the proof. THEOREM 6.5 If point D is in the interior of I ABC, then mlABC>mlABD and m£ABC> m£DBC. Proof: Since D is in the interior of /.ABQ ray BD is between rays BA andBC. By the Angle Measure Addition Postulate, mLABC = mlABD + m£DBC> Since mLABD and mLDBC are positive, it follows from Definition 6,3 that m Z ABC > m Z ABO and m Z ABC > m Z DBC. EXERCISES 6.2 In Exercises 1-10, identify the order property that is illustrated. L If AB < 6. then AB # a 2. If a - fe < 15 and ft < 3, then a < 18, 3. If x < 7 and 7 < y, then x < y. 252 Inequalities in Triangles Chapter 6 4. If a < 5, ihen 4o < 20. 5- If AB < RS and BC < ST. then AS + BC < HS + ST, 6. If Jm Z ABC > ^m Z RST, then m Z ABC> m £ R ST. 7. If Ali > CD and CD a EF, then AB > EF. 8. If x + 3 < 8, then a: < 5. 9. If x - y > 12 and y = 7, then x > 19. 10. If 3 > and 2 > 0, then 2 + 3 > 0. 11, in the figure, AD > BE and DC > EC. Prove AC >' BC. 12. Given the figure for Exercise 11, if m £ 1< m £2 and m Z3 < to Z4 prove that m Z ABC > m £ BAC. 13, Ii m/.A = 90 + k, where k > 0, and ZB is u supplement of ZA, prove that Z 2? is an acute angle. 14. In the figure, £D =z £ DEC. 15. In the figure, CD JL bis AB and Prove that m Z ABC > ffl Z D. C-P-B. Prove that AC > CE. 16. Prove part 2 of Theorem 6.1, 17. Prove part 2 of Theorem 6.2. 18. Prove Theorem 6.3. 19. Explain why Theorem 6.4 has the following consequence. If A, B, C are three distinct collinear points, then C is between A and B if and only if AC <A£ and BC< AB. 20. Explain why Theorem 6.5 has the following consequence. If D is a point in the interior of /ABC, then m I ABD < to £ ABC and m £DBC < m £ABC. P 21. In the figure, M is the midpoint of AT and BC. Prove tliul m£BCD> ml B. 6.2 Inequality for Number* 253 22. For each figure use your protractor to measure Z BCD y LA, and L B, Record your results in a table. How does m/ BCD compare with mLA in every cu.se? m Z BCD with m L B in every ease? m BCD m A mLB 254 Inequalities in Triangfes Chapter 6 23. challenge problem. Gi vet i the coplanar potato A,B,C, D, J#, P such that A, B, C are noncollincar, A~C~D, B-M-C, and A-M-P, prove that P is in the interior of L BCD. (Hint: You must show that P is on the tf-side of S/5 and that P is on the D-side of BC.) 24, challenge pbohlem. Prove the Transitive Property of Order (Prop- erty 0-3). {Hint: You will nocd to use Definitions 8.3 and 6.4 and the Addition Property of Kquality in your proof.) 6.3 THE EXTERIOR ANGLE THEOREM In both figures of Figure 6-1, ZABC, ZBGA, and LCAB are called interior angles of A ABC, We call /.BCD, which is adjacent to LACB and forms a linear pair with it, an exterior angle of A ABC. Both Z A and Z B are called nonadjacent interior angles of the exterior angle Z BCD. The term "interior angle" is convenient when you want to emphasize the distinction between an exterior angle of a triangle and an angle of a triangle. \ T ote diat the adjectives "adjacent" and "non- adjaeerit" apply to an interior angle and descrrt>e its relation to a par- ticular exterior angle. We formalin these ideas in die following definition. Definition 6.5 Each angle of a triangle is called an interior angle of the triangle. An angle which forms a linear pair with an interior angle of a triangle is ealled an exterior angle of the triangle. Each exterior angle is said to be adjacent to the in- terior angle with which it forms a linear pair and nonadjacent to the other two interior angles of the triangle. Every triangle has six exterior angles, two at each vertex, as shown in Figure 6-2, The two exterior angles at each vertex are vertical angles and hence are congruent. 6,3 The Exterior Angle Theorem 255 Figure fi-2 Note that L DCE in Figure 6-2 Is not an exterior angle. Why? In Figure 6-2 T L KAC and I HAB are the two exterior angles at vertex A of &ABQ and LABC and LBCA are the two nonadjacent interior angles of each of them. Name the two exterior angles at vertex B of A ABC in Figure 6-2 and their nonadjacent interior angles. If you worked Exercise 22 of Exercises 6,2, your results should sug- gest the following theorem. THEOREM 6.6 {The Exterior Angle Tfteorem) Each exterior angle of a triangle is greater than either of its nonadjacent interior angles. Proof: Let the vertices of the triangle be A . B, and C. T iet D he a point on AC such that C is be- n p tween A and D. (See Figure /S. ""/ 6-3.; We must prove that / X. LBCO> LBAC and that LRCD> LB. We first prove that j**- IBCDy LB. Figured Let M be the midpoint of BC and let P be the point on AM such that A-M-F and AM = MR Then AAMB == A PMC by S.A.S. (show this) and mLBCP = mLB. Why? Since P is in the interior of /BCD, m L BCD > m L BCP by Theorem 6.5. We have shown that mLBCD > mLBCP and that mLBCP- mLB. It follows from the Transitive Property (0-3) that mL BCD > mLB, and from Definition 6.2 we have LBCD > / B. 256 Inequalities In Triangles Chapters To prove that Z BCD > Z BAG, we use the midpoint of AC and show that the other exterior angle at C {LACE in Figure 6-4) is greater than / BAG in the same way as in the part above. Since the two ex- terior angles at C are congruent (Why?), it follows that IBCD> I BAG Figure <M The proof that we have given for one exterior angle of the triangle can be easily modified to show that die theorem holds for any of the six exterior angles of the triangle. However, it is not necessary to go to all this trouble since our choice of the exterior angle at C was stricdy an arbitrary choice. The fact that we were able to prove the theorem by choosing arbitrarily any one of the six exterior angles of A A /?C insures us that there is no need to prove the theorem for every exterior angle. In the proof of Theorem 6,6\ we stated that point P (in Figure 6^3) is in the interior of Z BCD. You are asked to prove this in the Exercises at the end of this section. COROLLARY 6.6, 1 If one of the angles of a triangle is a right angle, then the other two angles of the triangle are acute angles. Frvof: Assigned as an exercise. COROLLARY 6.6,2 If one of the angles of a triangle is an obtuse angle, then the other two angles of the triangle arc acute angles. Proof: Assigned as an exercise. It follows from the Exterior Angle Theorem (more directly from its corollaries, Corollary 6.0.1 and Corollary 6.6.2) that no triangle has more than one right angle or more than one obtuse angle. An important kind of a triangle is one that has one right angle. There are special names for triangles with a right angle and for triangles with an obtuse angle. 6.3 The Exterior Angle Theorem 257 Definition 6.6 A right triangle is a triangle with one right angle. The hypotenuse of a right triangle is the side opposite the right angle. The other two sides of a right triangle are called legs. Definition 6.7 An obtuse triangle is a triangle with one ob- tuse angle. Definition 6.8 An acute triangle is a triangle with three acute angles. Theorem 4.14 asserts that for each point on a line in a plane, there is one and only one line which (1) lies in the given plane. (2) contains the given point, and (3) is perpendicular to the given line. We can now use the Exterior Angle Theorem to prove a companion theorem. THEOREM 6, 7 Given a line and a point not on the line, there is one and only one line which contains the given point and which is perpendicular to the given line. Proof: Let I be the given line and P the given point not on L In part 1 of the proof we show that there is a line containing P and perpendic- ular to L In part 2 we show there cannot be two such lines. 1. Existence. (See Figure 6-5,) Let A and B be any two points of line I. Then PA is either perpendicular to / or not perpendicular to /. If FA ± h then the existence part of our proof is complete. If PA is not perpendicular to /, then there is a ray AC* with C on the opposite side of / from P such that I PAB == ABAC. Why? There is a point D on AC such that AD at Aft Why? P and C are on opposite sides of /, and C and D arc on the same side of l\ hence P and D are on opposite sides of /. Therefore PD intersects / at some point F. APAF^ &DAF by SJLS. (shew tills} and so L PEA as LDFA. Therefore, by Theorem 4.12, LPFA is a right angle and PF ± I Figure &S 258 Inequalities in Triangles Chapter 6 2. Uniqueness. (See Figure 6*6.) Let C be any point of / other than F and let H be any point of / such that G-P-H. Then Z PCFis an angle of APGF and is a nonadjacent * interior angle of the exterior angle Z PFIL Since m L PFH = 90, it follows from the Exterior Angle Theorem that mlPGF < 90 and therefore PG is not perpendicular to L It follows that PP is the only hue through P and perpendicular to L Figure fi-fi EXERCISES 6.3 1. Refer to Figure 6-4 and prove that Z BCD > Z MCj thus completing the proof of Theorem 6,6. 2. Prove Corollary 0.0. L 3. Prove Corollary 6.0.2. 4. Given A ABC with B-C-D and mLACD = 70, what must be true about m LABC? About r« Z BAG? About mAACB? 5. In the figure, name the two nonadjacent interior angles of ZBAP "Which exterior angle has A CAB and I ABC as its nonadjacent interior angles? Name all the exterior angles shown in the figure and their corresponding nonadjacent inte- rior angles. Using the figure, copy Exercises 6- 1 1 and replace the question marks by <, = t or > to make a true statement, a, b t c t x t y._ z denote angle measures. 6.3 The Exterior Angle Theorem 259 6. If b = 40, then x [?] 40 and y [?] 140, 7. If c = 90, then z \t] 90, a |Tj 90, b Q] 90, x (?)90, and y |7]90. 8. If a = 40 and b = 60, then 2 f7] GO. 9. If (/ = 140, then a [T] 140 and c [7| 140. 10. If x = 130 t then h [?] 130. 1L If a = 55 and c = 80, then i/ [?] 80. Refer to the figure for Exercises 12-18. In each exercise, arrange the nuni- bers in order, starting with the smallest. In these exercises, <i, h t c> d, e,f t g denote angles. 12. m La, m Z c 13. mLh t mLd 14. mle, mZc 15. mZb» BlZ/ 16. m £c ml_a t mLe 17. mZc, m Z e, rn Z g 18. mZf ( mZa,»nZg 1 »nZ« 19. In the Bgure, prove that mlACB > mLB. A 20. List the angles marked in the figure In Increasing order of size. 260 Inequalities in Triangles Chapter6 21 ♦ In the proof of Theorem 6,B it was stated that the point P (Figure was in the interior of L BCD, To prove this* wc must show that P is on tile D-sidc of S3 and on the B-side of c3. A C Copy and supply the missing reasons iu the following proof. Statement A-C-D, A-M-P, B-M-C M and F are on the same side of S3. M and B are on the same side of 53. P is on the B-side of CD. A and D are on opposite sides of BC. A and P are on opposite sides of BC, P is on the D-sidt of BC, P is in the interior of / BCD. L Given 2, Theorem 2.2 3. m 4. Statements 2 and 3 5. \?} 7, Statements 5 and 8. EKHHIl) 22. Is it possible for a triangle to have two right angles? Justify your answer. 23. Suppose that I is a line in plane a and that F is a point not in a. Docs Theorem 6.7 still apply? Draw a figure and explain your answer. 24. Given: AAMC f A-F-S> 25. Given: AADB.A-C-D, F-S-C AD = "KB Pmw,: Z1>Z2 Prove: LACW> LDBA 6.3 The Exterior Angle Theorem 261 h Cwtmt B is the midpoint of 333, 8 is the midpoint of FC, B-c-n Prom: IDCE> LA 27. Gtem: ARTS, R-W-l\ SW is the midray of L RST Prove; Z3>Z1 28, Given: Quadrilateral A BCD, B-K-C, 3D ss DE t L 1 ~ Z2 Prvce: Z3>Z1 29. Gkert: F is the midpoint of AD and BE Prove: Z2<Z1 30, Measure the sides in centimeters and the dingles in degrees of the scalene triangle in the figure. Record the measurements to the nearest tenth of a centimeter and the nearest degree in a table as shown. Yon will need to refer to the results of this exercise in Section 6.4. Angles c = f?l m/.A=\J\ ifiZB = f7| 262 Inequalities m Triangles Chapter 6 31. Draw three scalene triangles of different sizes and shapes and label the vertices and sides as in Exercise 30. Also make and record measure- ments for each triangle. You will need to refer to the results of this ex- ercise in Section 6,4. 32. a r allence problem , If Pis any point in the interior of A A 8C, prove that m£APB> mlC. 33. challenge pkohlem. Prove that the sum of the measures of any two angles of a triangle is less than ISO, 6.4 INEQUALITIES INVOLVING TRIANGLES If all three sides of a triangle are congruent, then die three angles of the triangle are congruent, and if two sides of a triangle arc congru- ent* then the angles opposite these sides are congruent. In this section we investigate how the angles of a triangle are related to each other when they are opposite Sides of unequal length. Refer to the table you prepared in Exercise 30 of Exercises 6.3 and answer the following questions. Which side is the longest? Which angle is the largest" How are the longest side and the largest angle of AABC situated with respect to each other? Which side is the shortest? Which angle is the smallest? How are the shortest side and the smallest angle of AABC situated with respect to each other? Observe the order relation among the measures of the angles of AABC and complete the following statement; mL\2] > mZQ] > mZ|T]. Observe the order relation among the measures of the sides of AABC and complete the following statement: EJ>E>[& How do the order relations among the measures of the angles of the triangle compare with the order relations among the measures of tile corresponding opposite sides of the triangle? Refer to the tables you prepared for the three triangles in Exercise 31 of Exercises 6,3 and answer the same questions as the preceding ones for each of these triangles. Make a general statement about the relative position of Lhe longest side and the largest angle of a triangle; the shortest side and the smallest angle. 6,4 Inequalities Involving Triangle* 263 The comparisons suggested by the preceding experiment are for- mulated in the following two theorems, THEOREM 6*8 {Angle-Comparison Theorem) If two sides of u triangle are not congruent, then the angles opposite them are not congruent and the greater angle lies opposite the greater side. Proof: Given; AABC, with AB > ~KC To From: I ACS > I ABC (See Figure 6-7.) Statement 1 „ AB > £C 2. AByAC 3. There is a point D on such that KD zzlC. 4. AD = AC 5. AB>AD 6. D is between A and B. 7. D is in the interior of £.ACB. 8. mlACB>mAACD 9. lACDss I ADC 10. mlACD = mlADC 11. nZACB > mZ40C 12. ZADC> ZAJ3C 13. m£ADC>mLABC 14. mZACB>mZABC 15. lACBy IABC Reason 1. Given 2. Definition 6. 1 3. Segment Construction The- orem 4. Definition of congruence for segments (3) 5. Substitution Property of Equality (2, 4) 6. Theorem 6,4 7. Theorem 4,11 8. Theorem 6.5 9. Isosceles Triangle Theorem (3) 10. Definition of congruence for angles (9) 11. Transitive Property (8, 10) 12. Exterior Angle Theorem 13. Definition 6.2 (12) 14. Transitive Property (U, 13) 15. Definition 6.2 (14) 264 Inequalities in Triages Chapter 6 We now state and prove the converse of Theorem 6.8. THEOREM 6.9 (Side-Comparison TJieorem) If two angles of a triangle are not congruent, then the sides opposite them are not congruent and the greater side lies opposite the greater angle, Proofs Let A ABC be any triangle with two angles that are not con- gruent Suppose it has been named so that these noncongruent angles are / B and Z C and such that LC> L B, In terms of this situation the hypothesis and the conclusion to be proved are as follows. Hypothesis: LC> LB Conclusion: AB > AC (See Figure 6-S.) Since AB and AC are numbers, one of the following must hold: (1) AB<AC (2) AB = AC (3) AB > AC Which property of numbers are we using here? The method of proof is to show that (1) and (2) are impossible, so (3) must hold, thus proving the theorem. (1) If AB < AC, then, by Theorem 6.8, LC < LB, This contra- dicts the hypothesis; hence AB < AC is impossible. (2) If AB = AC then A ABC is isosceles and LC ss LB. Again this contradicts the hypothesis; thus we see that AB = AC is impossible. It follows then that AB > AC so that AB > KC and the desired conclusion has been proved. The following two corollaries follow immediately from Theorem 6.9 and Corollary" 6.6.1. COR OLLAMY 6.9. 1 The hypotenuse of a right triangle is the longest side of the triangle. Proof: Assigned as an exercise. 6,4 Inequalities Involving Triangles 265 COROLLARY 6.9,2 The shortest segment from a point to a line not containing the point is the segment perpendicular to the line. restatkmeoti Given a Hue I and a point P that is not on I, if PA J. I at A and B is any other point of ?, then PA < FB. (See Fig- ure 6-9.) ►J Proof: Assigned as an exercise. a B Figure 6-9 When we speak of the distance between a point P and ft line l> we naturally mean the shortest distance from J' to I. It follows from Corol- lary 6,9.2 that there is such a shortest distance, and so we make the following definition. Definition 6,9 The distance between a point and a line not containing the point is the length of the perpendicular seg- ment joining the point to the line. The distance between a line and a point on the line is defined to be zero. It is customary to associate three "distances between ft point and a line" with every triangle. With A ABC there is associated the distance between A and EC, the distance between B and CA, and the distance between C and AS. Any side (or its length) of a triangle may be thought of as the base. Associated with each base is the segment (or its length) joining the opposite vertex to a point of the line containing the base. The following definition has two parts. In (I), we define base and altitude, thought of as segments. In (2), we define base and altitude, thought of as numbers (lengths of segments or distances). Definition 6,10 L Any side of a triangle is a base of that triangle. Given a base of a triangle, the segment joining the Opposite vertex to a point of the line containing its base, and perpendicular to the line containing the base, is the altitude corresponding to tluit base. %. The length of any side of a triangle is a base of that tri- angle. The distance between the opposite vertex and the line containing that side is the corresponding altitude- 266 Inequalities in Triangle* Chapter 6 Figure 6-10 shows two triangles, A ABC and AA'B'C. In each tri- angle the segments from the vertices to the lines containing the oppo- site sides have been drawn. These segments are the altitudes of the triangles. Thus XD is an altitude of A ABC. It is the altitude from A to sideTJC. The point D is the foot of the altitude from A to EC. Note for an acute triangle, such as AABC in the figure, that the foot of each altitude is an interior point of a side. Note for an obtuse triangle, such as A A'B'C in the figure, that the feet of two altitudes ar« • not points of the triangle. Even though point E\ for example, is not a point of side A'C. it is frequendy called the foot of the altitude from B' to side A^U. Figure 640 Draw a right triangle, A ABC, with the right angle at C. Draw the altitude from C to the hypotenuse. Is the foot of this altitude a point of the triangle? Name the other two altitudes of A ABC. Are the feet of these altitudes points of the triangle? In Chapter 3 we postulated that if A, B, C are noncolliuear points, then for distances in any system, AB -\- BC > AC. The postulate was called the Triangle Inequality Postulate. We now have the necessary geometric properties to prove this statement as a theorem. THEOREM 6.10 (Triangle Inequality Theorem) The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Proof: Given any triangle, it follows from the Trichotomy Property for numbers that there is one side of the triangle which is at least as long as each of the other two sides. Suppose, in A ABC, that BC j> AB that BC > AC, (See Figure 6-11.) Figure 6-11 B 6.4 Inequalities Involving Triangles 267 We must prove the following three statements: (1) AB + BC>AC (2) AC +BOAB (3) AB + AC > BC (1) By hypothesis, BC > AC. Since AB > T (that is, AB is a positive number), we have AB + BC>AC by the Addition Property of Order (0-4). (2) By hypothesis, BC > AB. Since AC>O t we have AC + BC > AB. Why? (3) On opp AC choose point D so thai AD = AB. (See Figure 6-12.) C Figure G-12 Since A is between C and IX A is in the interior of L DBC. Also mLDBOmlABD by Theorem 65. But r»ZABD = mLADB by the Isosceles Triangle Theorem, and so mlDBOmlADB. "Therefore, by the Side-Comparison Theorem (applied to A DBC), DC>BC Since DC = DA + AC (Why?) and DA = AB, we have DC = AB 4- AC We have shown that DC > BC and that DC = AB + AC. Therefore, by the Substitution Property, AM + AC>BC It follows from Tneorem 6.10 that, in informal language, the short- est distance between two points is the length of the segment joining them. In formal geometry, of course, once a unit of distance is given, there is only one distance between two points. It follows from Theorems 6.8 and 6.9 that, in any one triangle, the greater angle lies opposite the greater side and conversely, the greater side lies opposite the greater angle. Let us now consider two compan- ion theorems involving two triangles. 268 Inequalities In Triangles Chapter 6 In Figure 6-13 t AB = A'W and EC = BC. What can you say about these triangles if m LB=mL B'? How do AC and A 'C compare? B a* Fitfirt-6-13 A C A' In Figure 6-14, we again have AB = AB' and BC = B'C\ but this time m L B > m L B\ Is the correspondence ABC < — *■ A'BC/ a con- gruence? I Tow do AC and A'O compare in this case? B B' A A' Flip j re 6-14 In Figure 6-15, we have AB = A f B\ m£B<C.ni£W t Ts the correspondence ABC * — i ence? How do AC and A'C compare in this case? B' BC-BC 7 and A'BC a congru- A' c Figure 615 A On the basis of the preceding discussion, it would appear that if two sides of one triangle are congruent, respectively, to two sides of a second triangle and if the corresponding included angles are not con- gruent v then the sides opposite these included angles are not congru- ent and the side opposite the larger angle is the larger side. It may help you to understand this last statement by examining a pair of compasses. Observe that, as the angle formed by the pointers of the compasses gets larger, BO does the distance between the ends of the pointers. We make this idea formal with our next theorem. THEOREM 6.11 (Side-Comparison Theorem for Tivo Triangjtes) If two sides of one triangle are congruent, respectively , to two sides of a second triangle, and if the angle included by the sides of the first triangle is greater than the angle included by the sides of the second triangle, dieu the third side of the first triangle is greater the third side of the second triangle. 6,4 Inequalities Involving Triangles 269 restatement: Given A ABC and AA'B 'C w ith AB — A*ff and BCsrFC. If Z B> LB\ then aC>~FU. (See Figure 6-16.) B Hr C Figure 6-16 Proof: By the Angle Construction Theorem, there is a point (J> on the *— > Oside of AB such that ZAB(<) = ZA'JJ'C", (See Figure 6-17.) Since A ABC > ZA'FC, P is in the interior of /.ABC, Choose point P on such that BF -- There are now three cases to consider. Cose J. P is in the exterior of A ABC, Case 2, P is on AC Qwf' 3. F is in the interior of AABC, We consider only Case 1 here- The proofs of Cases 2 arid 3 are left as exercises. Proof of Case h In Figure 6-17, AABP s A A 'B'U by S.A.S. Let BR he the midray of Z CBP intersecting AC at £. Then APBE ss ACB£ by S.A.S. Applying the Triangle Inequality Theorem to AAEP, we have AE + EF> AP. Figure 6-17 But AP a A'C (Why?) and £P = EC. Why? By the Substitution Property, we get AE + EC > A'C, and by the Distance Betweenness Postulate, AC > A'C. Therefore T^C > /TTT and this completes the proof for Case 1. 270 Inequalities in Triangles Chapter 6 The converse of Theorem 6.1 1 is also true and we state it as our last theorem of this section. THEOREM §.12 (Anglo-Comparison Theorem for Two Triangles) If two sides of one triangle are congruent, respectively, to two sides of a second triangle, and if the third side of the first triangle is greater than the third side of the second triangle, then the angle in- cluded by the two sides of the first triangle is greater than the angle i iK- hided by the two sides of the second triangle, restatement: Given A ABC and AA'B'C with A~B ^ A^W and BC^&C, if ZC>A 7 &> then /_B> IB . (See Figure 6-18.) Figure &I8 Proof: Since m Z B an d nt Z B' are numbers, one of the following must hold: (1) mlB<mlB' (2) mlB=mlB' (3) ml.B>mLW Which property of numbers are we using here? The method of proof is to show that (I) and (2) are impossible, so (3) must hold* and the theorem will be proved. The problem of showing that (!) and (2) are impossible is left as an exercise. EXERCISES 6.4 t. Given A ABC with AZ? = 12, BC = 15, and AC - 10. name the. angles in order of size beginning with the angle of least measure. 2, In A $KM S mlS = 47, mlK = 85 t andml M = 48. Name the short- est side; the longest side. a Name the longest side of A ABC if (a) ml A = 44, mlB = 90; (b) ml A = 120, mlB = 40. 4, Does a triangle exist with the following numbers as side lengths? Why? (a) Si 3, 10 (b) 5, 3, 8 (ci 5, 3, 4 5. Given A ABC with A-B-D, if ml ABC > mlCBIX prove AC > BC 6.4 Inequalities Involving Triangles 271 6. Given the following figure with angle measures as marked, for each of the three triangles name the sides of the t dangle in order of increasing length 7. In the figure for Exercise 6, which segment is the shortest? 8. Given the following figure with angle measures as marked, prove that CD is the longest segment. 9. In the figure, if the angles have the indicated measures, which segment is shortest? In Exercises 10-12, got the "best" answer you can in the sense that the smaller number is as large as possible and the larger number is as small as possible, 10. Copy and complete: If the lengths of two sides of a triangle arc 7 and 12, then the length of the third side must be greater than Q] and less than [?]. 11. Copy and complete: If the lengths of two sides of a triangle are 6 and 9, then the third side must have a length less than JTJ and greater than [T|, 12. Between what two numbers must the length of the third side of a tri- angle lie if two of its sides have lengths of 17 and 28? 272 Irwqualtties in Triangles Chapter 6 13, Given A-C-D, E-C-B, and with angle measures as marked, prove that HE > AD. 14. Prove that the sum of the lengths of the diagonals of a convex quadri- lateral is less than the perimeter of the quadrilateral 15, Given AABC with D a point on AC such that HD bisects /.ABC, prove that AB > AD. 16, In the figure, FS < SA and PQ < QR. Prove that m£SPQ>m£SRQ. 17, Prove Corollary 6,9.1. (Hint: You will need to use Corollary 6.6.1 and Theorem 6.9 in your proof.) 18, Prove Corollary 6.9.2. (See Figure 6.9.) Given AF > PB, prove m L AC? > m L RCF. {Hint: Use Theorem 6.12.) Given AFQR with M die midpoint of TQ, if mlRMQ>m/ PMR, prove that @B > ER, 6,4 Inequalities Involving Triangles D 'C 273 21. Given a convex quadrilateral ABCD with AD = BC, if AB > DC com- pare m Z CAD with m Z BCA. 22. Prove the following theorem. THEOREM TIa Y b,c are the side lengths of a triangle and if h a is the altitude corresponding to base a, then h a < h and /i a < c. 23. chai.i.fnce problem. Prove Case 2 of Theorem 6.11. This is the Case whore point P is on SC, (Hint: In Figure fi-i7 for Theorem 8.11, recall that AABP=z A A' JVC You must show that AC > A'C.1 24. challenge problem. Prove Case 3 of Theorem 6.11. Tins is the Case where point P is in the interior of A ABC. (Hint: Let BR he the bisector ray of ACBP, intersecting AC at E. Why is Ai^E a ACBE? Why is A£ + BC > A/»? Prove /££ > 5^?.) 25. challenge problem* Complete the proof of Theorem 6. 12 by show- i tig tliat Cases 1 and 2 are impossible. (Hint: Use Theorem 6. 1 1 for Case 1 and the S.A.S. Postulate for Case I.] CHALLENGE PROBLEM. If P is any point in the interior of A ABC,, prove that AP + PB<AC + CB. (Hint: Let AP intersect W at D. Apply the Triangle Inequality Theorem to AACD and to ABDF,) 274 Inequalities in Triangles Chapter 6 CHAPTER SUMMARY In this chapter we deed I with geometric inequalities involving angles and sides for any one triangle and also for two triangles. We defined GREATER THAN and LESS THAN for angles in terms of their measures and for segments in terms of their lengths. We stated some order properties for numbers which are listed below by name only. You should lenow the complete statement of each of these properties. 0-1 POSITIVE CLOSURE PROPERTY 6-2 TRICHOTOMY PROPERTY 0-3 TRANSITIVE PROPERTY 0-4 ADDITION PROPERTY 0-5 MULTIPLICATION PROPERTY The key theorem concerning geometric inequalities is the EXTERIOR ANGLE THEOREM which states that an exterior angle of a triangle is greater than either of its two nonadjaeent interior angles. Other important theorems involving inequalities in any one triangle arc listed below by name only, ft is important that you be able to state these theorems in your own words and that you understand their proofs. THEOREM 6.8 (77u? Angle Comparison Theorem) THEOREM 6,9 (The Side-Comparison Theorem) THEOREM 6.10 (Tlie Triangle Inequality Theorem) Tile following two theorems give inequalities concerning two triangles. Be sure that you know the complete statement of these theorems. THEOREM 6.11 (The Side-Comparison Theorem for Two Triangles) THEOREM 6T2 (27k? Angle-Comparison Theorem for Two Triangles) We defined the DISTANCE between a point and a line not containing the point to be the long til of the perpendicular segment joining the point to the line. The distance between a line and a point on the line is defined to be zero. We proved that the hypotenuse of a right triangle is the longest side of the triangle and that the shortest segment from a point not on a line to the Hue is the segment perpendicular to the line. Rovltw Exercises 275 REVIEW EXERCISES In Exercises 1-7. identify the order property that Is illustrated, 1. If m/ A>mZBandmZ.B = mZC, theu mLA>m£C. 2. If x - y > 5 and y = 4 t then x > & 3. If m£A> mlB, then JtwZA > |»»Zfl. 4. If i- /- y, then % > y or * < y, 5. If A£ < ES and BC < S£ then AB + BC < RS + ST, 6. If i + i/ > z, then x > s — r/. 7. If o + fe > c and a * /> < d t then d > c. 8. In the figure, what must he true about tnAD? About mZ /JF£? AbouL I •:-: 2) J? Refer to Figure 6-19 for Exercises 9-14. Copy each exercise and replace the question marks by the symbol < f =, or > to make a true statement. 9, h Q] tl 10. ag]c|T]e 11. cQd 12. c[T|/ 13. fl|?J/ 14. b^]d[T\f\2\c\2XQ 15. Which theorem do the markings on the figure contradict? Figure 6-19 16. In the figure, which angle is the largest? The smallest? 276 Inequalities in Triangles 17, In the figure, which side is the longest? The shortest? Chapter 6 18. In the figure, if the angles have the indicated measures, which segment is shortest? What theorem did you use to decide? 19, In the figure, if the ingjtea have the indicated measures, arrange the seg- ments PQ, QR, rT5, ?7S, PS, undTTi in order of increasing length. 20, Copy and complete th«- fi blowing statement, putting the largest number in (a) and the smallest number in (b) which will make a true statement If the lengths of two sides of a triangle are 8 and 15. then the length of the third side must be greater than (a) (?] and less than (b) [T]- Review Exercises 277 Exercises 21-26 refer to Figure 6-20 showing A ABC with A-D-C. B-E-Q B-F-D, and A-F-E, You should be able to defend your answers using the- orems that you know. Figure 6-20 21. Name five angles whose measures arc less than m L 10. 22. Name two angles whose measures are less than mZ.5. 23. Name four angles whose measures are greater than mL 1. 24. L 7 is an exterior uugje of which two triangles? 25. Is /_ 6 an exterior angle of ABFE? 26. Is ml& > mZ2? Explain why. 27. In the figure, if BC is the longest side and 7%\ is the shortest side, prove that m/,A > m£C, (Hint: Draw GA and use the Angle-Comparison Theorem.) 28. Given: SK = KM ftere KM > KJ [Hint; Use the Kxterior Angle Theorem, the Isosceles Triangle Theorem, and the Side-Comparison Theorem.) 278 Inequalities in Triangles 29. For A ABC prove that AH - WO < AC, Chapter 6 30. If, In AAjBC, BC = AC, prove that AC > JAB. 31. Given th e fig ure with angle and side measures as marked, name the sides SR t RQ t and QPin order of increasing length. State a theorem that justifies your conclusions. 32. Given the figure with sides us marked, copy and insert < or > in (a), (l>), and (c) so that each is a true statement. (a) roZAJTJmZBEC (b) mZDEC[?]mZBEC (c) ittZJi 0mZDEC 33. Slate theorems that justify your conclusions in Exercise 32, R*vl*w exercises 279 34. Given: &ABC with median AM mLAMB = 70 Prove; mABymlC CMS 35, If ASOf is a right triangle with S-M-C, prove that KC > KM K 36. challenge problem. If A, B, C, D are four distinct points in space and if no three of these points lie on a line, prove that AB + EC + CD> DA. 37. challenge PROBLEM. Prove that in any triangle there are two sides of lengths r and s such that ±<t<2. Dat;kl Phnctltm/Phuto Ke$eurclwr,\ Parallelism 7.1 INTRODUCTION We are now ready to consider one of the most basic ideas in geom- etry, the idea of parallel lines. What do yon think of when someone says "parallel lines?" Perhaps the lines which separate lanes on a straight running track or the cracks between the boards in a floor, or the strings on a violin? What are some of the properties of parallel lines? If two ships are sailing parallel courses close to one another as sug- gested in Figure 7-1 , what is the relationship of the angles marked in the figure? Figure 7-1 282 Parallelism Chapter 7 The congruence of these angles is an example of an Idea from in- formal geometry that suggests a property of parallel lines in formal geometry. The set of rails on a railroad track must fit wheels which arc fixed so that the distance between them cannot change as they roll down the track. This suggests another property of parallel lines in formal georn- eLry, the property of being the same distance apart everywhere. What arc the basic properties of parallel lines? Can you think of one or more properties such that if lines have these properties they should then be considered to be parallel lines? It would be natural to use such basic properties in deciding on a definition for parallel lines. ITie most basic properties of points, lines, and planes are the Inci- dence Properties which were discussed in Chapter L Two different lines either intersect or they do not, IF they do not intersect and are copianar, we call them parallel lines. If they do not intersect and are not copianar, we call them skew lines. We know that two distinct in- tersecting lines determine a plane. It follows that, if two lines are not copianar, then they must he nonintersecting lines and hence are skew lines. We shall find it convenient to consider a line as parallel to itself. We begin our formal treatment with several definitions based on the foregoing ideas. 7.2 DEFINITIONS Definition 7.1 Two distinct lines which are copianar and nonintersecting are parallel lines, and each is said to be paral- lel to the other. Also, a line is parallel to itself. The lines in a set of lines are said to be parallel lines if each two in the set are parallel. Definition 7.2 TWo lines which do not lie in the same plane are called skew lines. \ citation. If p and q are lines, then p | q means that p is parallel to (f, and pX° means that p is not parallel to q. Note that every pair of distinct parallel lines arc copianar, that every pair of distinct intersecting lines are copianar, and that every pair of skew lines are noncoplanar. If p and q are lines, there are four distinct possibilities: 7.2 Definitions 283 1. p and q are noncoplanar, in which case they axe skt^u . 2. p and q are coplanar but not parallel, in which case they inter- sect in exactly one point and there is exactly one plane which contains them. 3. p and q arc distinct parallel lines, in which case they do not in- tersect and there is exactly one plane which contains them. 4. p and q are nondistJnct parallel lines, in which case p and q are the same line and there are infinitely many different planes con- taining p and q t EXERCISES 7.2 Exercises 1-5 pertain to the rectangular box suggested in Figure 7*2. E Figure 7-£ 1. Using A, H, C, , . ■ , designate two distinct edges which lie ad parallel lines. 2. Using A, B, C, . , , , designate two distinct edges which lie on intersect- ing lines. 3. Using A, J3 T C, . , , , designate two distinct edges which lie on skew lines. 4 How many edges of the box lie on lines which are parallel to AB? 5. How many edges of the box lie on lines which are skew to TiB? 6. If p and q are skew lines, why is there no plane which contains both? 7. If p is a line in a plane a t how many different lines in a are parallel top? 8. If p is a line in a plane a t how many different lines in a are perpendicular top? 0. If p is a line in a plane a, how many different lines in a are skew to p? 10, Prove that two distinct parallel lines detenninc exactly one plane. 11, Given that m is a line and P is a point on tit, prove that there is one and only one line through P and parallel to RL 12, Let m be a line in a plane a and P a point in a but not on wi. Use your knowledge regarding the existence of perpendicular lines and the Ex- terior Angle Theorem for triangles to prove that there is at least one line through P and parallel to m. 284 Parallelism Chapter 7 7.3 EXISTENCE OF PARALLEL LINES Answering questions regarding existence is an important part of building a formal geometry. How do we know that there are such things as poinLs, lines, and planes? We know that they exist because of the postulates and the theorems of Chapter I on incidence relations. Given two distinct points A and B on a line l t how do we know that there is a point Con/ such that A-C-B and AC - CB? We know it liecause we can prove it! Our proof depends in a very essential way on the Ruler Postulate, Given a line A B in a plane a, how do we know that there exists a ray AC in a such that m/.RAC = 30? We know it be- cause we can prove itl Of course, the proof is easy once we adopt die Protractor Postulate- Given a line I and a point P not on I, how do we know that there exists at least one line through P and parallel to J? It is true in some ge- ometries that there are no parallel lines. But in our geometry t the ge- ometry that we inherited from Euclid, there are parallel lines, and furthermore, we can prove it. In fact, you were asked to prove it in Exercises 11 and 12 of Exercises 7*2, These exercises arc combined in the next theorem. THEOREM 7.1 (Existence of Parallel Lines Theorem) If I is a line and P is a point, then there is at least one line through P and parallel to I. If P is on I, Hi ere is exactly one line through P and paral- lel to i Proof: There are two cases to consider: (1) P is on I and (2) P is not on I Case 1. We suppose first that P is on if. Since I is parallel to I, it follows that there is a line through P and parallel to L If m is any line different from I and through P t then it intersects I in exactly one point and hence is not parallel to L Therefore there is one and only one line through P and parallel to I. Case 2. Suppose next that P is not on I (See Figure 7-3.) Then P and / determine a plane. Why? Gall It a. 4 ll Figure 7-3 7.3 Existence of Parallel Lines In a there is a unique line m through P and perpendicular to I (see Theorem 6.7) and a unique line n through P and perpendicular to m. We shall prove that n and / are parallel. Suppose, contrary to what we want to prove, that n and / are not parallel. Then m, 11, and / are distinct intersecting lines forming a tri- angle with an exterior angle and a nonadjaccnt interior angle both of which are right angles. (See Figure 7-4.) But this is impossible in view of the Exterior Angle Theorem. It follows that n and I arc parallel. Therefore there is at least one Hue through Panel parallel to I This com- pletes the proof. 285 Figure 7-4 Along with questions regarding existence in mathematics there are sometimes questions regarding uniqueness. Theorem 7.1 settles the matter of existence of parallel lines, but it docs not settle the matter of uniqueness. If / is a line and .Pis a point on /, we know that there is one and only one line through F and parallel to L If / is a line and P is a point nof on I, we do not know yet that there is one and only one line through P and parallel to L For about 2<XX) years following the time of Euclid, mathematicians tried to prove that, given a line and a point not on the line, there is a unique line through the given point and parallel to the given line. Finally two mathematicians, a Russian named Nikolai Ivanovitch Lobachevsky (1793-1856) and a Hungarian named Janos Bolyai (1802-1860), proved independently that it is impossible using only the postulates of Euclid (other than his parallel postulate) to prove the uniqueness of parallels. Since we want uniqueness of parallels, we follow hi the footsteps of Euclid and adopt a "parallel postulate," We defer the statement of lliis postulate, however, to Section 7,8. In Sections 7.4 and 7.5 we introduce the concept of a transversal and develop some theorems whose proofs do not depend on the Parallel Postulate. These arc theorems that belong both to the ordinary geom- etry of Euclid and to the non-Euclidean geometry of Lobachevsky and Bolyai in which the Euclidean Parallel Postulate is replaced by a postu- late which says that parallels are not unique. More specifically, the Bolyai-Lobachevsky Postulate states that if Ms any line and P is any point not on 1, dien there are at least two distinct lines through P and parallel to I 28« Parallelism Chapter 7 7.4 TRANSVERSALS AND ASSOCIATED ANGLES Let p, (/, and f be three distinct lines in a plane a. The line t may in- tersect both p and q or it may not If t intersects both p and q, then it may intersect them in different points or it may intersect them in the same point In Figure 7-5, p and q are intersecting lines and t intersects p and q in different points. In Figure 7-G, p and q are parallel lines and t intersects p and q in different points. In Figure 7-7, t intersects both p and q in the same point. In Figure 7-8, t intersects q but does not intersect p. Figure 745 P + < + 7 Figure 7-7 Figure 7-S In situations like those in Figures 7-5 and 7-6 we say that f is a trunsc&rsal of p and q. In each of these figures t intersects the union of p and 9 in a set consisting of two distinct points. On the other hand, in Figures 7-7 and 7-8, t docs not intersect the union of p and q in a set consisting of two distinct points. As we said, in Figure 7-5, / is a trans- versal of p and q. In Figure 7-5 q is a transversal of p and f, and p is a transversal of q and f. We arc now ready for the following formal definition. 7.4 Transversals and Associated Angles 287 Definition 7.3 A transversal of two distinct coplanar lines is a line which intersects their union in exactly two distinct points. In Figure 7-9, A s B t C, D are four of the vertices of a rectangular box. li Figure 7-9 Note that AB, BC, CD are three distinct lines, BC intersects the union of .4 B and CD in two distinct points. Name them. is not a transversal of AB and CD. Why not? Is it true that if a line is a transversal of two other lines, then the three lines arc distinct eoplanar lines? Give a reason for your answer, A transversal of two lines forms with these lines eight distinct an- gles. For convenience we gjve names to certain pairs of these angles. Figure 7-10 In Figure 7-10, * is a transversal of p and c/; angles 1, 2 t 3, 4 are the angles formed by p and i; angles 5, 6, 7, 8 are the angles formed by q and t. Angles 1, 4, 6, 7 are called ulterior angles and angles 2, 3, 5, 8 are called exterior angles , 2*K taniliiin er7 Angles 1 and 6 are one pair of consecutive interior angles. Figure 7-1 1 shows these angles with several points labeled. Notice that Z 1 is the union of BA and ED and that Z 1 is LABD. Express Z 6 in terms of rays and write a name for it involving names of points. Figure 7-11 9 Notice that Z 1 and Z 6 have a segment in common and that their interiors are on the same side of the transversal. Name the segment that is die intersection of these two angles. Lines p and q t which we are considering in connection with a trans- versal f, might be parallel or they might not be. if p and q are not paral- lel, then they intersect in a point. This point might be on the opposite side of the transversal from the interiors of a pair of consecutive in- terior angles, as it would be for Z 1 and Z 6 in Figure 7*11, or it might be on the same side, as it is for the pair of consecutive interior angles 4 and 7 in Figure 7-12. Note that although the intersection of Z 1 and Z 6 is a segment, the intersection of Z4 and Z 7 is the union of a seg- ment ami a set consisting of a single point. Name that segment and that point. Thus the intersection of two consecutive interior angles may be a segment or it may be the union of a segment and a set consisting of a single point. Figure 7-12 We shall give a formal definition of consecutive interior angles after we have introduced two other phrases for angles associated with u pair of Uiies and a transversal. 7.4 Transversals and Associated Angles 289 In Figure 7-13, Z 1 and L 7 are interior angles but not consecutive interior angles and not adjacent angles. We call them alternate interior angles. The intersection of Z 1 and Z 7 is a segment and their interiors are on opposite sides of the transversal. Figure 7-13 Another pair of alternate interior angles are Z 4 and Z 6, (See Fig- ure 7-14.) Their intersection is a segment and their interiors lie on op- posite sides of the transversal. Figure 7-14 In Figure 7-15, Z I and Z 6 are consecutive interior angles. Angles 1 and 2 are adjacent angles. Angles J. and 7 are alternate interior angles. This brings us to Z 1 and £5. Figure 7-15 290 Parallelism Chapter 7 Angle 1 is an interior angle and Z 5 is not. Their interiors lie on the same side of the transversa]. In Figure 7-16, these angles are shown with several points labeled. Examine the intersection of Z 1 and £5. Figure 7-ia Is the intersection of these angles a ray s the ray BP? Observe Lhat the intersection of die interiors of Z 1 and Z 5 is the interior of Z 5. Angles 1 and 5 are called corresponding angles. Another pair of corresponding angles associated with the lines p and q and their transversal t are Z 4 and Z 8 as shown in Figure 7-17. The intersection of these angles is the union of a ray and a set consist- ing of a single point. The interiors of these angles lie on the same side of the transversal. Although neither interior contains the other, they do intersect. Figure 7-17 Thus far in this section we have introduced alternate interior an- gles, consecutive interior angles, and corresponding angles in connec- tion with two coplanar lines and a transversal. Actually, it is not neces- sary to refer to these lines and the transversal in describing these angles. Indeed, no such reference is made in the following definitions. The phrase "a segment and a point*' appears in the following defini- tions. We accept these phrases as a short way of saying "die union of two sets of points, one of them a segment and the other a set which consists of a single point/' We accept "a my and a point** to mean "the union of two sets, one of them a ray and the other a set consisting of a single point." 7.4 Transversals and Associated Angles 291 Definition 7.4 Two coplanar angles are alternate interior angles if their intersection is a segment and if their interiors do not intersect. Definition 7.5 Two coplanar angles are consecutive interior angles if their intersection is a segment, or a segment and a point, and if their interiors intersect. Definition 7.6 Two coplanar angles are corresponding an- sxlcs if their intersection is a ray, or a ray and a point, and if their interiors intersect. EXERCISES 7,1 1, How do yon know that parallel lines exist in Euclidean geometry? Is your reason a postulate or a theorem or u definition? & If p is a line, if q is a line, and if p is parallel to q, is it possible that the intersection of p and q is a set which contains more than one point? Explain. 3. If p is a line, if q f§ a line, and if the intersection of p and q is the null set, is it possible that p and q are not parallel? Explain. 4. If p is a line, if q is u line, and if p is parallel to q, is it possible that there is no plane containing the union of p and q? Explain, ■ Figure 7-18 sh o ws t wo coplanar lines p a nd q a nd a transversal t with several points labeled. Copy Exercises 5- 10 and replace the question marks with one word,, two words, or three capital letters so that the resulting statement is a true sentence concerning Kijjyre 7-18. Figure 5* LBDC and ZQ arc corresponding angles, fl, / BDC and Z[7] are alternate interior angles. 7. L BDC and Z |T| arc consecutive interior angles, a L FDC and LAQD are \?J angles. 9. / HDF and Z HOP. are CD angles. 10, Z FDC and / ECD arc [T| angles. 292 Parallelism Chapter 7 Figure 7- 19 shows two copbnur lines p and q and a transversal t. Eight an- gles are labeled. Copy Exercises 1 1 -20 and replace the question marks with one word or two words or a letter or a number so as to make a true statement about Figure 7-19, BrgHK 7-19 11. a and C are vertical [?]. 1& a and d are \T\ angles. 13. a and u are [?] angles, 14. c and c are [7| angles. 15. c and [7] are corresponding angles. 16\ x and [f| are corresponding angles, 17. d and [?] are alternate interior angles. 18. There are JT| pairs of alternate interior angles among the eight angles. 19. Thereare {7] pairs of consecutive interior angles among the eight angles. 20. There are \T\ pairs of corresponding angles among the eight angles. <■ — ■» < — * «. — t Figiue 7-20 shows three iioncollineur points A, B f C, the lines AB, HC, CA, and 12 angles formed by than, Copy Exercises 21-23 and replace the ques- tion marks so as to make a true .statement about the figure. Figure 7-20 21. / 1 and ]7] are corresponding angles; also / 1 and [Tj are corresponding angles. 22. Z 12 and \T\ are alternate i ntcrior angles; also L 12 and [?] are alternate interior angles. 7,4 Transversals and Associated Angles 293 23. _ 1 2 and IT| are consecutive interior angles; also / 12 aud [T] are con- secutive interior angles, Figure 7-21 shows two distinct parallel lines, a transversal, and eight associ- ated angles. Copy Exercises 24-28 and replace the question marks so as to make a true statement about the figure. "4- Figure 7-21 24, Angles 1 and 5 arc corresponding angles; their intersection is Q]; I heir interiors lie on [7] of L 25, Angles 3 and 5 are consecutive interior angles; their intersection is [Tj: their interiors lie on [7] of t, 26, Angles 3 and 6 are alternate interior angles; their intersection is Q]; their interiors lie on [7] of f , 27* Angles 6 and 7 are vertical angles; their union is the [tj of a pair of lines; their intersection is Q]; the intersection of their interiors is [7], 28. Angles 5 and ft are a linear pair of angles; their union is the union of a line and a [T|; their intersection is JT]; the union of these angles and their interiors is the union of a halfplane and Q], 20, The figure shows A ABC and two rays in the same plane such that B is between A and E r and BD is between liE and EC. Considering only angles that can be named in terms of three points labeled in the figure, identify all pairs of alternate interior angles, 30. Under the same conditions as those in Exercise 29„ identify ail pairs of consecutive interior angles, 31. Under the same conditions as those in Kxercise 29, identify all pairs of corresponding angles. 294 ParalJefism Chapter 7 Figure 7-22 shows a quadrilateral and one of its diagonals. In Exercises 32-35, copy and complete each statement so that it will be true. 32. / ACH and Z CAD are alternate interior angles detennmed by trans- versal aS and lines \f\ and [7]. 33. Z BAC and / DC A are alternate interior angles determined by trans- versal |TJ and lines \T\ and {T}, 34 Z ABC and Z BCD arc \T} angles determined bv transversal jT] and lines (U and Q] . 35. Z BAD and Z arc consecutive interior angles detennined by trans- versal [T\ and lines and [7). Figure 7-23 is a plane figure with five angles labeled, including two pairs of vertical angles. Let m Z a m 65 and mZa = m / e. In Exercises 36-38 t find the measure of the given angle 36. Li 37. Zu 38. Ze Figure 7-23 Figure 7-24 is a plane figure showing two lines and a transversal. Refer to tills figure for Exercises 39-42. Figure 7-24 7.5 Some Parallel Line Theor&m*, 295 39. Name (a) the pairs of alternate interior angles; (b) tlie pairs of consecutive interior angles; (c) the pairs of corresponding angles. 40. Prove: If one pair of alternate interior angles are congruent, then (a) the other pair of alternate interior angles are congruent; (b) each pair of consecutive interior angles are supplementary: (c) each pair of corresponding angles arc congruent 41. Prove: If one pair of consecutive interior angles are supple mentary, then (a) the other pair of consecutive interior angles are supplementary; (b) each pair of alternate interior angles arc congruent; (c) each pair of corresponding angles are eongruent 42. Prove: If one pair of corresponding angles are congruent, then (a) each pair of corresponding angles are congruent; (b) each pair of alternate interior angles arc congruent; (c) each pair of consecutive interior angles are supplementary, 7.5 SOME PARALLEL LINE THEOREMS This section contains several theorems that are sometimes helpful in proving that two lines are parallel. THEOREM 7,2 Let two distinct coplanar lines and a transversal be given. If the transversal is perpendicular to both lines, then the lines are parallel, Proof: Let a and b be distinct coplanar lines and let t be a transversal of them. (See Figure 7-25.) We wish to prove that if a and h are per- pendicular to t t then a and b are parallel. Suppose, contrary to this as- sertion and as suggested by the figure, that a _L t. that feif, and that a is not parallel to b. Then a and b intersect in some point P forming APVQ. This triangle has Z 1 as an exterior angle and Z 2 as a nonadja- cent interior angle. By hypothesis, both of these angles arc right angles, so &F¥Q is a triangle with one exterior angle congruent to a nonadja- cent interior angle. 296 Parallelism Chapter 7 According to the Exterior Angle Theorem, however, the measure of an exterior angle of a triangle is greater than the measure of cither nonadjacent interior angle- 1 truce we have two angles which are con- gruent by one line of reasoning and which are not congruent by an- other line of reasoning. We arrived at this contradiction after we had supposed that die lines a and h are not parallel Since we cannot have a contradiction in our system, we have proved that lines a and h cannot lie not parallel. Hence they are parallel This completes the proof of the theorem. We pause briefly to comment on this proof. It is an example of an indirect proof, The theorems in our formal geometry arc true state- ments. Their truth rests on a foundation of postulates and definitions. In proving theorems, that is, in establishing their truth, we may use definitions, postulates, and theorems proved previously. flow did we prove Theorem 7i2? Consider the given situation in- volving lines a T h r and t. Lines a and h are distinct and coplanar, and / Is perpendicular to both of them. We do not know that a and h are par- allel nor do we know that a and b arc not parallel But we do know from our definitions that a and b are either parallel or they are not par- allel Since one of these possibilities leads to a contradiction, we con- clude that the other possibility is the valid conclusion. It may be helpful to see this reasoning in skeleton form using sym- bols. We start with a situation in which a and b are coplanar lines and t is a transversal of them. Let P and Q stand for statements as follows: P: t ±a and t 1 b. Q. tffc Then our theorem and proof arc essentially as follows. THEOREM If P, then 0. Proof: It is given that P is true. Either Q is true or Q is false. If QU false, then it follows that Ph false, This contradicts the hypothesis that P is true, Therefore Q is true, and the proof is complete. If P denotes a statement, it is convenient to denote the opposite statement by not-H If P is a true statement, then not-F is a false state- ment If P is a false statement, then not-P is a true statement. If a state- ment has the form "If P, then Q" then there is always a related state- ment of the forrn "If not-Q* men not-JV' This related statement is called the contrapositivc of the given statement. If the contrapositive of a statement is true, then the statement is true. For, if not-Q implies not-P, then the only way it is possible for J* to be true is for {) also to be 7.5 Some Parallel Line Theorems 297 true. In other words, if not-Q implies not-P, then P implies Q. This means that we can prove a theorem by proving its contrapositive. If it is easier to prove the eontrapositive, then this is what we should do. A proof using the contrapositive is one form of indirect proof. For example, suppose we wish to prove if a 2 =f=b 2 s then a^=h. An easy way to prove this is to prove its conlrapositive if a = h, then a 2 = h 2 . Tlie proofs of the following three theorems are assigned as exercises. THEOREM 7,3 {Alternate Interior Angle Theorem) If two al- ternate interior angles determined by two distinct coplanar lines and a transversal are congruent then the lines are parallel. THEOREM 7.4 (Corresponding Angle Theorem) If two corre- sponding angles determined by two distinct coplanar lines and a transversal are congruent, men the lines are parallel. THEOREM 7.5 [Comecutwe Interior Angle Theorem) If two consecutive interior angles determined by two distinct coplanar lines and a transversal are supplementary, then the lines are parallel EXERCTSES 7.5 Figure 7-26 shows u transversal of Iwo distinct coplanar lines and eight as- sociated angles, In Exercises 1—4, measures of two of the angles are given. Explain why m must lie parallel to n. m + ««- Figure 7-26 1. m/.a = 105, m£e = 105 3. mLc = 75, mLv = 105 2. mid = 105, mZf= 105 4. m t a = 75, m L f = 75 29S Parallelism Chapter 7 Figure 7-27 shows five coplanar segments and several associated angles. In Exercises 5-9, state which lines must be parallel on the basis of the given information. 5. Z«s Iv 6. Zpss lq 7. ml ADC + mlDCB = 180 8. ml ADC + ml BAD = ISO 9. ml DAB + ml ABC = 180 Figure 7-28 shows five distinct coplanar points A, B, Q D, E. Points A, B, C are coHinear, AD 1 Tkff, and CE _L AC In Exercises 10-14, state whether this given information implies the Stated conclusion. 10. A3 | AC 11. A~$ | M 12. a6 if ci 13. AD | EC 14. XBl\ BE In Exercises 15-20, the figures show three coplanar lines. The measures of three angles, expressed in terms of a number x, are given in the figures- Find x and determine whether m is parallel to n. Give a reason for your answer. 15. 16. 7,5 Some Parallel Una Thtortms 299 18. m4- "4- 2x + 5 ™«- lOO + i/ 80 + * *4- 100 + 21 21. Given: A plane figure with AB = CD, AD = BC. Trove: AD \ B&. XS jj DC SOD Parallelism Chapter 7 22. Given: A figure with A-D-B, A-E-Q AD = A£ AB = AC. LACBz* IADE Prove: DE\BC 23. Given: A figure in which A~C and SD bisect each other at M, Prove: A~B \ S3, A/5 || S3 2? C 24. Gitcn. A figure with A-R-B, B-P-C, C-Q-A. AQ -QC= RP CF = PBzzQR RR = RA = PQ Prove: m/_A + m£B + m I C = 180 25. Git en. A plane figure, with I A =z IB, AD = BC, A£ = EB t DF=m Prove: EF 1 W, AB J EF, DC || A~S U Lil £ III .C A £ rt 26. Prove Theorem 7.3. 27. Prove Theorem 7,4. 28. Prove Theorem 7.5. 29. Write the contraposiuve of Theorem 7,3. Is this coiitrapositive a true statement? 30. Write the contrapositive of Theorem 7.4. Is this contrapositive a true statement? 31. Write the contrapositive of Theorem 7.5. Is this contrapositive a true statement? 7.6 The Parallel Postulate 301 7.6 THE PARALLEL POSTULATE AND SOME THEOREMS Theorem 7. 1 states that through a given point not on a given line there is at least one line parallel to the given line. As we pointed out following the proof of that theorem, it is impossible, using only Pos- tulates 1-25, to prove that this parallel line is unique. Since we want it to be unique, we adopt the following Parallel Postulate. As we stated in Section 7.3 it is this postulate which makes our geometry, Euclidean geometry, different from that of Lobachevsky and Bolyai. POSTULATE 26 (Parallel Postulate) There is at most one line parallel to a gjven line and containing a given point not on the given line. This postulate and Theorem 7.1 tell us that if a line and a point not on it are given t then there is exactly one line through the given point and parallel to the given line. Furthermore, as we said in Section 7.3, we know that if a line and a point on it are given, then there is exactly one line through the given point and parallel to the given line, namely the given line itself. Hence, in every case, given a line and a point, there is exactly one line through the given point and parallel to the given line* In Section 7.5 there are some parallel line theorems, actually the- orems stating conditions that imply that lines arc parallel. These the- orems are useful in proving lines parallel. In this section we have the converses of several of these theorems, These converses were deferred until now because the Parallel Postulate is essential for their proof. If a theorem is of the form "If P, then Q," then its converse is the associated statement "If Q, then ?" It should be clear that the con- verses of some true statements are not true statements. For example, 'Tf a number is greater than 100, then it is greater than 6" is a true statement, whereas its converse, c *If a number is greater than 6, then it is greater than 100," is certainly a false statement The converses of some theorems are theorems: the converses of other theorems are not theorems. We shall prove that the converses of Theorems 7.3, 7.4, and 7.5 are theorems. Note first, however, that the converse of Theorem 7.2 is not true. To see this, suppose m and n are distinct parallel lines and that f is a transversal of them which makes an augle of measure 30 with m. Then, according to Theorem 7.7 given later in this section, it also makes an angle of measure 30 with line n. Thus wc see that lines m and n may be parallel even though a transversal is not perpendicular to both of them. We proceed now to the converses of the other theorems in Section 7.5. 302 Parallelism Chapter 7 THEOREM 7.6 (Converse of Alternate Interior Angle Tlieorem) Tf two distinct lines are parallel then any two alternate interior angles determined by a transversal of the lines are congruent. Proof: (See Figure 7-29.) Let m and n be distinct parallel lines. Let t be a transversal that intersects them in P and Q, respectively. Let R and S be points on m and n, respectively, such that R and S are on op- posite sides of t. Wc shall prove that LRPQ == LSQT. t Figure 7-W It follows from the Angle Construction Theorem that there is a ray ~PR\ with W on the tf-side of U such that LR'PQ as Z 5QK Since Z RTQ and Z SpP are congruent alternate interior angles, it follows from Theorem 7.3 that lines AT and n are parallel. Therefore AT and m are lines through P and parallel to n. From the Parallel Postulate, however, it follows that there is only one line through F and parallel to n. Therefore iiT and m are the same line. Then Z R'PQ and Z RPQ are the same angle and it follows that Z RPQ m Z SQP. THEOREM 7, 7 (Converse of Corresponding Angle Theorem) Tf two distinct lines arc parallel, then any two corresponding angles determined by a transversal of the hues are congruent. Proof; Assigned as an exercise. THEOREM 7,8 (Converse of Consecutive Interior Angle The- orem) If two distinct lines are parallel, then any two consecutive interior angles determined by a transversal of die lines are supplementary. Proof: Assigned as an exercise. As we noted above, the converse of Theorem 7.2 is not true. It should be noted, however, that a theorem similar to Theorem 7.2 does have a true converse. This is Theorem 7.9. Its converse is Theorem 7.10. Tlieorem 7.9 follows immediately from Theorem 73. Theorem 7.6 The Parallel Postulate 303 7.2 begins with a "situation statement" (Let two distinct coplanar lines and a transversal he given) followed by an "If P, then Q'* type of .state- ment In Theorem 7.9 some of the P has been put into the "situation/' but the meaning of the two sentences taken logcdier is the same as in Theorem 7.2. Theorem 7.10 follows immediately from Theorem 7.7. THEOREM 7.9 I^et a and h be two distinct coplanar lines, and let t be a transversal of them that is perpendicular to a. If t is per- pendicular to b, the lines a and b are parallel. THEOREM 7.10 Let a and h l>e two distinct coplanar lines, and let f he a transversal of them that is perpendicular to a. If a and b are parallel, then f is perpendicular to h. The next theorem provides anodicr useful method for proving lines parallel, TTTEOR EM 7,11 Two coplanar lines parallel to the same line are parallel to each other. Proof: Let p and q be coplanar lines each parallel to a line r. If p — q t then p is certainly parallel to q. Suppose, then, that p and q are distinct lines. There are two possibilities as indicated in Figure 7-30. Either p and q are parallel lines or else they have exactly one point T say P, in 14- P'oeelbUity (») Powubility fb) Figure 7-30 common. If they have exactly one point P in common, then there arc two distinct lines through P and parallel to r. Since this contradicts the Parallel Postulate, it follows that possibility (h) in Figure 7-30 is impos- sible, and therefore p is parallel to q. If we consider only lines lying in a given plane, we see that the re- lationship of parallelism for lines is an equivalence relation. That is.. It is reflexive, symmetric, and transitive, (1) It is reflexive since every line is parallel to itself. (2) It is symmetric since if line p is parallel to line </, then line q is parallel to line p, (3) To show that it is transitive, suppose that line p is parallel to line r and that line r is parallel to line q. Then, since parallelism in a plane is symmetric, it follows that line q is par- allel to hue r. Then p and q are both parallel to line r and it follows from 304 Parallelism Chapter 7 Theorem 7,11 that p is parallel to q. Later we shall see that parallelism for lines is an equivalence relation even without the restriction that the lines all lie in one plane. THEOREM 7A2 Let three distinct eoplanar lines with two of them parallel be given. If the third line intersects one of the two parallel lines, then it intersects the other also, Proof: Assigned as an exercise. THEOREM 7.13 Let two sets % and 3 of parallel lines in a plane q he given, (This means that every two lines in § are parallel and that every two lines in 3 are parallel.) If one line in § is perpendic- ular to one line in 3 , then every line in § is perpendicular to every line in 3. Proof: Let s be a line in S and let t be a line in 3 such that s is per- pendicular to t. Let u be any line in g and let v be any line in 3 . We want to prove that u is perpendicular to c. Suppose, first, that s and u are distinct lines and that r and i" are distinct lines as indicated in Fig- ure 7-31. Since * intersects t and i is parallel to t% it foDows from The- orem 7-12 that 8 intersects v and therefore s is a transversal of the par- allel lines t and o. Then it follows from Theorem 7.10 mat s is perpendicular to v. Similarly, it follows from Theorems 7.12 and 7.10 thai c is a transversal of the parallel lines * and u and that v is perpen- dicular to u. This completes the proof for the case in which s and u are distinct lines and t and are distinct lines. i ► 8 - u« ► Figure 7-31 Suppose next that s = u and that s is perpendicular to t. Then it follows as above (really the halfway point in the reasoning of the pre- ceding paragraph) that s is perpendicular to v. Tlie ease in which s and u are distinct and t = c is assigned as an exercise. The case in which s = u and I = v requires no proof since the hypothesis and the conclusion are the same in this instance. 7.6 The Parallel Postulate 305 EXERCISES 7,6 1. Use Theorem 7,6 to prove Theorem 7 J. 2. Use Theorem 7.6 to prove Theorem 7.8. 3. Prove Theorem 7.7 without using Theorem 7.6, (This proof will be sim- ilar to I he one for Theorem 7.6 that uses the Parallel Postulate.) 4 Prove Theorem 7.8. (This proof will be similar to the one for Theorem 7.6 that uses the Parallel Postulate.) 5. Explain (he following statement; Of the three theorems, 7,fi, 7,7, 7.8 t any one of them may be considered as the basic theorem and the other two as corollaries of it. 6. Prove Theorem 7.13 for the case in which a =5^ u and t = c. In Figure 7-32, lines o, b, c, and d are eoplanar and the measures of several angles are marked. In Exercises 7-9, justify the given assertion regarding the lines in this figure. Figure 7:12 7, a is parallel to h. 8. a is parallel to c. 9* h is parallel to c. 10. JSa t b,c are eoplanar lines such that a is parallel to b and b is parallel to c+ does it follow that a is parallel to e? Is this an instance of the reflexive property of parallelism for lines? Of the symmetric property? Of the transitive property? 11* If a t b, are eoplanar lines such that a is parallel to b and b is parallel to c, docs it follow thai b is parallel to n? Is this an instance of the reflexive property of parallelism for lines? Of the symmetric property? Of the transitive property? 12. If a, h. c are eoplanar lines such that a is parallel to b and b is parallel to C does it follow that c is parallel to c? Is tilts an instance of the reflexive property of parallelism for lines? Of the symmetric property? Of lite transitive property? 306 Paralii - Chapter 7 In Figure 7-33, a, b t c, d are coplanar parallel lines, f is a transversal of ihcm, and the measures of se%'eral angles are marked. In Exercises 13-18, justify the given assertion regarding angle measure. 13. x= 117 14. y = 63 15. z - 117 16. u - 63 17. o = 63 18. w = 63 af b + t + d + J Figure 7-3.1 in F.xerciscs 19-21, use Figure 7-34 with E-A-D, m Z EAB = 1 IS, KD | BC. 19. Find™ lABDHZm LABD = 3m Z.DBC. 20. Find m /DBC and in ZBDA if 2m lABD = 3m I DBC 22. Given a plane figure made up of two parallel lines, a transversal, and a segment, with angle measures as marked, find x\ y, and z. 23. Given a plane figure made up of two pairs of parallel lines, with angle measures as marked, find x, y, z, u, v, and to. 7.6 The Parallel Postulate 307 24. Given a plane figure with AB DE. with A. C, E all on the same side of BD, with C on the fc-side of AB, and with C on the A -side of ED, prove that m Z BCD = m / ABC + m Z COE (Hint- Copy the Figure and draw the line through C parallel to 12.) 4 — *■ n 25, In the figure for Exercise 24, what is the sum of the measures of Z ARC, LCBD, LBDC&ad /LCDE? 26, In the figure for Exercise 24, what is the sum of the measures of Z CUD* ZBDC. and ZDCB? 27, The figure helow shows some segments that are parallel and some that are not parallel. Try to write a good definition of parallel segments. Punt"!,-: Parallel Not pimiltet V Not paralM 25. The figure below shows some quadrilaterals that are parallelograms and some that are not. Try to write a good definition of parallelogram. 308 Parallelism : :: - ~ 7.7 PARALLELISM FOR SEGMENTS; PARALLELOGRAMS In this section we introduce some definitions and theorems con- cerning an important type of quadrilateral called a parallelogram. Since the sides of a quadrilateral and, hence, of a parallelogram are segments., we need a formal definition that extends the concept of parallelism to segments. In Section 7,8 we extend the concept also to rays. Definition 7.1 If the lines which contain two segments are parallel, then the segments are said to lie parallel segments. and each is said to be parallel to the other. The segments in a set of segments are parallel if every two of them are parallel. The lines which contain parallel segments need not be distinct. In other words, a segment of a line is parallel to every segment of that line. As a special case of a special case, we note that every segment is parallel to itself. Let A, B, C t D be four points with A^B and C ^ D. {See Figure 7-35.) Then KB is parallel to TJD if and only if AS is parallel to CD. We use the same symbol to denote parallelism for segments that we use for lines. Thus AB . CD means that AB is parallel to CD. C D -t ^i ► Figure 7-35 Definition 7.8 A parallelogram is a quadrilateral each of whose sides is parallel to the side opposite it Consider a parallelogram ABCD as shown in Figure 7-36. Since ABCD is a quadrilateral, it follows that AB, BC, CD, DA are four dis- tinct lines- Since An and CD are distinct parallel lines, it follows that C and D He on the same side of AB* Similarly, D and A lie on the same side of BC, A and B lie on the same side of CD, and B and C lie on the same side of DA. Therefore each side of a parallelogram lies on a line which is the edge of a halfplane that contains all of the parallelogram D C Figure 7-36 7.7 Pararffelitm for Segments; Parattetograrm 309 except that side. Therefore every parallelogram is a convex quadri- lateral. (The word convex is used here in the sense of a convex polygon. See Definition 4, 1.6.) The next three theorems state some important properties of parallelograms. THEOREM 7.14 If a convex quadrilateral is a parallelogram, then its opposite sides are eongnient. Proof: Let parallelogram ABCD be^ given. (See Figure 7-37.) Then AB i| CD and BC \\ DA. We shall prove that AB = CD and that BC = DA. First draw the seg- ment A€, Our plan is to use con- gment triangles, Statement 1, ZBACs IDCA % ACzsZA 3. AACBsa LOAD 4. AAiJCss &CDA 5. ab ^ £E anc i bc ^ DA Reason 1. Alternate interior angles de- termined by a transversal of two parallel lines are congruent, 2. Why? 3. Why? 4. Why? 5. Why? JHEOREM 7.15 If a convex quadrilateral is a parallelogram, then its opposite angles are congruent. Proof: Assigned as an exercise. THEOREM 7, 16 If a convex quadrilateral is a parallelogram, then its diagonals bisect each other. Proof: Assigned as an exercise. The next three theorems are useful in proving that certain quadri- laterals are parallelograms. THEOREM 7.17 If two sides of a convex quadrilateral are parallel and congruent, then the quadrilateral is a parallelogram. Proof: Assigned as an exercise. 310 Parallelism Chapter 7 TITEORFM 7J8 If the diagonals of a convex quadri lateral bisect each other, then the quadrilateral is a parallelogram. Proof: Assigned as an exercise. THEOREM 7.19 If each two opposite sides of a convex quadri- latend are congruent, then the quadrilateral is a parallelogram. Proof: Assigned as an exercise. Just as parallelograms are convex quadrilaterals with a special prop- erty, so are trapezoids convex quadrilaterals with a special prop- erty (only not quite as special as that for parallelograms). Definition 7.9 A trapezoid is a convex quadrilateral with least two parallel sides. Note that we do not say two and only two sides parallel. A trape- zoid may have only one pair of parallel sides or it may have two pairs of parallel sides. In other words, every parallelogram is a trapezoid, but not every trapezoid is a parallelogram. In some l*x>ks trapezoids are restricted to have only two parallel sides. In these instances the set of all trapezoids and the set of all parallelograms do not intersect. In this book the set of all parallelograms is a subset of the set of all trapezoids. Rhombuses, rectangles, and squares are all parallelograms with special properties. Their formal definitions come next. Definition 7.10 A rhombus is a parallelogram with two ad- jacent sides congruent Definition 7.11 A rectangle is a parallelogram with a right angle. Definition 7.12 A square is a rectangle with two adjacent sides congruent. THEOREM 7.20 A rhombus is an equilateral parallelogram. Proof Assigned as an exercise. 7,7 ParalltHsm tor Segments; ParallefogTams 311 THEOREM 7.21 A rectangle is a parallelogram with four congru- ent angles. Proof: Assigned as an exercise. THEOREM 7.22 A square is an equilateral rectangle. Proof; Assigned as an exercise, THEOREM 7,23 A square is an equiangular rhombus. Proof: Assigned as an exercise. THEOREM 7.24 The diagonals of a rhombus are perpendicular. Proof: Assigned as an exercise. In order to prove that a figure is a rhombus it is sufficient to prove that it is a parallelogram with two adjacent sides that are congruent (Definition 7.10). Tf we know that a figure is a rhombus, then we may conclude that all four of its sides are congruent (Theorem 7.20), If we want to show that a parallelogram is a rectangle, it is sufficient to show that it has one right angle, since it then necessarily has four right angles (Definition 7,11 and Theorem 7,21' lb show that a rect&ngfc is a square, it is sufficient to show that two of its adjacent sides are congru- ent, since then all four of its sides are congruent (Definition 7.12 and Theorem 7.22}. To show that a rhombus is a square, it is sufficient to show that it has a right angle since it is easy to show that an equilateral parallelogram with a right angle is an equilateral rectangle, that is, a sqiiare. In Chapter 3 we introduced the concept of distance. We restricted ourselves there to the idea of the distance between two points. We are ready now to extend the idea oi' distance to the distance between two parallel lines. Of course, we agree that the distance between a line and itself is zero. So let us consider the idea of the distance between two distinct parallel lines m and n as suggested in Figure 7-38. The length of a segment (sec the "dashed" segment in the figure) joining a point of one line to a point of the other line might be very long depending on bow the endpoiuLs are picked. « i v^ 1 1 312 Parallelism Chapter 7 It seems natural to think of the distance between two parallel lines as the length of the shortest segment joining the two lines, and it seems that this segment should be perpendicular to both lines. If we pick any point on m, say M, then there is a line t in the plane of m and n that is perpendicular to m at M , (See Figure 7-39.) Also I intersects n in some point* say A 7 , and MN is perpendicular to both m and n. If t' is another transversal of m and n perpendicular to both of them and intersecting them in M' and -V, respectively, how do we know that M.N and M'N' are congruent? Our idea for the distance between two parallel lines will not be any good unless we can show that MN and M'S 7 are con- gruent. The next theorem shows that this is possible. Figure 7-39 THEOREM 7.25 For every two distinct parallel lines there is a number that is the common length of all segments perpendicular to both of die given lines and with one end point on one of the given lines and one eiidpotnl on the other one. Proof: {See Figure 7-40.) I>et m and n be two distinct parallel lines. a b Figure 7-40 Let A and B be two distinct points of m, and C and D two distinct points of n such that the segments AC and BD are perpendicular to both m and n. Then m is a transversal of AC and BD and is perpendicu- lar to both. It follows that KC and WD are parallel (Theorem 7.2); hence ABCD is a parallelogram (by definition) and AC = BD (Theorem 7.14), If we think of segment KC in Figure 7-40 as a fixed segment and BD m a variable segment (one that we can pick anywhere as long as B is on m, D is on n, and ¥D is perpendicular to m and to n), then the num- ber we are looking for, the one whose existence wc wanted to prove, is the number AC. The idea of Theorem 7,25 is sometimes expressed by saying that two parallel lines arc everywhere equidistant. We are now ready for the following definition. 7.7 Parallelism for Segments; Parallelograms 313 Definition 7.13 The distance between two distinct parallel lines is the length of a segment which is perpendicular to both lines and whose endpoints lie on these lines, one endpoint on one line and the other endpoint on the other line. The dis- tance between a line and itself is zero. EXERCISES 7.7 1. Frove: If the convex quadrilateral ABCD is a parallelogram, then Z A fix Z C and I B as ID. (This is Theorem 7.15.) 2. Prove: If the convex quadrilateral ABCD Is a parallelogram, then A~C and WD bisect each other. (This is Theorem 7.16.) 3. Frove: II ABCD is a convex quadrilateral, if AB m CH5, and if AB || CD, then ABCD is a parallelogram. (This is Theorem 7.17.) 4. From: If ABCD is a convex quadrilateral, and if AC and #D bisect each other, then ABCD is a parallelogram. (This is Theorem 7.18,) 5. Frove: If ABCD is a convex quadrilateral if AB ==? CD and All as BC t then ABCD is a parallelogram. (This is Theorem 7.19.) 6. If ABCD is a rhombus and if AB =s AD, then AB = BC = CD = DA and ABCD is an equilateral parallelogram. {This is Theorem 7.20.) 7. Prove Theorem 7.21. 9. Prove Theorem 7.23. 8. Prove Theorem 7.22. 10. Prove Theorem 7.24, 11. Why are opposite sides of a parallelogram parallel? 12. Why are opposite sides of a parallelogram congruent? 13. The figure below shows some rays that are parallel and some that are not parallel, Try to write a good definition of parallel rays. A B c • > 4 • • » AB and BC are parotid Parallel rays Not parallel rays Not parallel nyt 14. The figure below shows some antiparallel rays and some rays that are not antiparallel. Try to write a good definition of antiparallel rays. A B CD — • ► 4 — • — • • m BA and EC are antiparalle] * ?Cot antiparaU*! BA and CD are antipartllel B ACnndAB are not aariparalM 4- Antiparallr-l my* Not antiparaCe] 314 Parallelism Chapter 7 7,8 PARALLELISM FOR RAYS In this section wo extend the concept of parallelism to rays. Since segments are parallel if the lines containing them arc parallel, we might consider rays parallel if the lines containing them are parallel. But it is useful to Distinguish between the case suggested by rays AB and CD and the case suggested by rays EF and GH in Figure 7-41. In both cases the rays lie on parallel lines. In one case, however, they point in the same direction, whereas in the other case they point in opposite directions. (We use the word "direction" here in an intuitive sense; it is not a part of our formal development*) Figure 7-41 Definition 7A4 (Sec Figure 7-42.) Two noneollincar rays AB and CD are parallel if AB and CD are parallel lines and if B and D lie on the same side of AC. Two coDmear rays are parallel if one of them is a subset of the other. Definition 7,15 (See Figure 7-42.) Two noneollincar rays EF and GH arc anliparallel if EF and GH are parallel lines and if F and H lie on opposite sides of EG, Two collinear rays are an ti parallel if neither is a subset of the other. AB unci CD urv porulkl. Figure 7-42 IJ*n±JK JIutAIKm iirti pontile]. EF mul GH are antiparalkL The proofs of tile following three theorems are assigned as ex- ercises. THEOREM 7.26 If Z A BC and I DEF are coplanar angles with BA and eB parallel and b8&xuI EF* parallel, then Z ABC^ Z DEF. 7.8 Paraltellsm for Rays 315 THEOREM 7.27 If Z ABC and t DEE are coplanar angles with — * — * ■ — ^ — * BA and ED parallel and with BC and Eb antiparallel, then / ABC and LDT.E are supplementary. THEOREM 7.2S If / ABC and Z OFF am coplanar angles with BA and ED antiparallel and with BC and EF antiparallel then LABC^ A DEE EXERCISES 7.S Dra%v a line and on it mark five points A, B, C, D, £ in the order named. In Exercises 1-20, slate whether the given .statement is true or fake. 2. AB = 52 3. 31 =s M 4. AB = M 5, AB || BA 6. AB || AC 7, IS || CD &.2Sc3ft 9. AB c BC 10, a2 C AC ilaScbS 12. 5? C AB 13, Ai! and BA arc parallel. 14* An and BA are anti parallel. 13, AB and FD are parallel. 16. HC and BA are antiparallel. 17. b? n bI = {B} 18. Bf? f| DA = BD 19. BC? P DA = Wi 20. d£ 1CD= A t B T C, . . . , K, L are distinct coplanar points, as in Figure 7-43, so that t j -^ ) < | ^ j DF | j/i and D/f | Fi. In Exercises 21-30, state whether the given rays are parallel or antiparallel, 21. 15i and 77? 22. DE and 77? 23. £// and BE 24. I1A and Li 25. ^andEF 26. CF and FC 27. CF andlf 28. Cpmidjf 29. Gland 77f 30. GfmidjH Hfure7-43 316 Parallelism Chapter 7 ■ In Exercises 31-36, copy and complete the statements. 31. The intcrseetitm of two parallel collinear rays is [?]. 32. The intersection of two parallel noncollinear rays is [7], 33. The intersection of two antiparallel collinear rays is (?] or [Tj or (?]. 34. The union of two parallel collinear rays is (t). 35. The union of two anti parallel collinear rays is a line with the interior points of a segment deleted or it is an entire fT|. 36. If ABVD is a parallelogram, then An and CD are [7] rays. 37. Prove Theorem 7.26 for the case in which BA and ED are collinear and parallel and BC and JSFftrc noncollinear and parallel. (See the figure be- low.) The case in winch BA and ED are noncollinear and parallel and BC and EF are collinear and parallel can be proved in a similar way. 38. Prove Theorem 7.26 for the case in which BA and ED are noncollinear and parallel and BC and £F arc noncollinear and parallel as suggested in the figure. (Hint-. Think of the figure formed by the lines Afi, BC, DE. and EF. Do you see that this figure contains a parallelogram? Then use Theorem 7.15.) 39. In Exercises 37 and 38, you were asked to prove Theorem 7.26 except for the case in which BA and ED are collinear and parallel and BC and EFare collinear and parallel. Draw a figure for this special case. Note that in this special case the theorem amounts to saying that equal angles are congruent angles. 40. Draw an appropriate figure and prove Theorem 7.27 for the case in which BA and ED are collinear and parallel and BC and EF are collinear and antiparallel. 4L Draw an appropriate figure and prove Theorem "7.27 for the case in which BA and ED are noncollinear and parallel and BC and EF are collinear and antiparallel, 7.8 Parallelism for Rays 317 42. Draw an appropriate figure and prove Theorem 7,27 for the case in which BA and ED are collinear and par&tlel and BC and EF are non- coUinear and antiparallel. 43. Draw an appropriate figure and prove Theorem 7.27 for the ease ia which BA and ED are noncollmear and parallel and BC and £F are noo- coDinear and an tj parallel, 44. Prove Theorem 7.28. Do it by eases. In which special case does the as- sertion of the theorem amount to an assertion that vertical angles arc congruent? 45. ABCD is a convex quadrilateral. Prove: If m£A +m/.B + mlC + m£D = 360, ifmZA a mZ C, andif mZB = roZ D t thenABCDis a parallelogram. 46. Prove: If Z A and Z B are consecutive angles of a parallel ©gram, then they are supplementary. (How many pairs of consecutive angles does a parallelogram have?) In Kxercises 47-52, write a sentence justifying the truth of tho given statement. 47. If two alternate interior angles determ ined by a transversal of two given copianar lines are not congruent, then the given lines are not parallel. 48. If two corresponding angles detennined by a transversal of two given copianar lines are not congruent, then the given lines are not parallel. 49. If two consecutive interior angles determined by a transversal of two given copianar lines are not supplementary, then the given lines are not parallel. 50. If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. 51. If two opposite sides of a quadrilateral are not parallel, then the quadri- lateral is not a parallelogram. 52. If ABCD Is a parallelogram, then AB and CD are everywhere equidistant. 53. Given two alternate interior angles detennined by a transversal of two parallel lines, prove that the angle bisectors of these angles are antiparallel ■ 54. 'Hie figure shows five copianar lines and some (perhaps all} of their points of intersection. If m^L ABE + mLBEA + mLEAB = 180, I BAE at LEAC, Z ABE b Z DBE t and m Z BEA = 90, prove that BD is parallel to AC. 318 Parallelism Chapter 7 7.9 SOME THEOREMS ON TRIANGLES AND QUADRILATERALS One of the most important consequences of die Parallel Postulate is the following theorem. THEOREM 7«$9 The sum of the measures of tlie angles of a tri- angle is 180. Proof* (See Figure 7-44,) Let A ABC be given. Let m be the unique line through C and parallel to AB. Let D and E be points on m such that D-C-E and such that B and D are on opposite sides of AC as in the figure. Then mLA + mLB + mlACB = mlACD + m L BCE + mZ ACB = 1 SO. (Which theorems justify these equations?) It may be of interest to mention here that in the non-Euclidean ge- ometry of Lobachevsky and Bolyai the sum of the measures of the an- gles of a triangle is less than 180, The sum is not the same number for all triangles in that geometry. The smaller the triangle is the closer die sum is to 180. THEOREM 7,30 The measure of an exterior angle of a triangle is equal to the sum of the measures of its nonacljaeent interior angles. Proof: Assigned as an exercise, THEOREM 7,31 Let a one-to-one correspondence between die vertices of two triangles be given. If two angles of one triangle are congruent, respectively, to the corresponding angle* of the other triangle, then the third angles of the two triangles are also congruent. 7.9 Theorems oti Triangles and Quadrilaterals 319 Assigned as an exercise- (Note Figure 7-45 which suggests that the correspondence need not be a congruence. One of the triangles may be larger than the other.) Figure 7-45 THEOREM 7.32 {The S.AJi. TJieorem) Let a one-to-one cor- respondence between the vertices of two triangles be given. If two angles and a side opposite one of them in one triangle are congru- ent, respectively, to the corresponding parts of the second triangle, then the correspondence is a congruence. Proof: Assigned as an exercise. THEOREM 7.33 The sum of die measures of the angles of a con- vex quadrilateral is 360. Proof: Let ABCD be a convex quadrilateral as in Figure 7-46. Draw diagonal BD. Since ABCD is a convex quadrilateral, D is in the interior of Z ABC and B is in the interior of Z CDA. Then, using the notation of the figure, we have mLA + mZABC + mLC + mLCDA = mZA -r {mil + m/2) + m£C+(mZ4 + mZ3) = (m IA + m Z 1 + m Z3) + m Z2 + (m LC + m Z4) = 180 + 180 = 360, THEOREM 7.34 The acute complementary. es of a right e are Proof: Assigned as an exercise. 320 Parallelism Chapter 7 THEOREM 7.-35 (The Hypotenuse-Leg Theorem) Let there be a one-to-one correspondence between the vertices of two right triangles in which the vertices of the rigfrt angles correspond. If the hypotenuse and a leg of one triangle are congruent to the corre- sponding parts of the other triangle, then the correspondence is a congruence. Proof: (See Figure 7-47.) Let A ABC with mZA = 90 and ADEF with m£_D = 90 be given. Let it be given that BC^EF and AB = DE. We shall prove that ABC * — > DEFis a congruence. You will be asked in Exercise 19 below to supply the missing reasons, tatement 1. Let G be a point on opp DF such that DG s AC, '2, mLEDG = 90 3. ED = BA and ml. A 4. ADEG =s AABC 5. EG^BC 6. BC = EF 7. EG a EF 8. ZEGDa ZKFD 9. AOEG a ADEF 10. AABCss ADEF - <.K) Reason 1. The Segment Construction Theorem 2. Z £Z>G is a supplement of a right angle 3. Given 4. S.A.S. 5. a 6. Given 7.Q] 8. E 9. S.A.A. 0. The equivalence properties of congruence for triangles EXERCISES 7.9 In Exercises 1-6, the measures of two angles of a triangle are given. Find the measure of the third angle of that triangle, 1. 93 and 80, % a and b. (Express the answer in terms of a and h) 3. 25 and x. (Express the answer in terms of x.) 4, 90 + k and 90 — 2k, (Express the answer in terms of Jt.) 7.9 Ttworems on Triangles and Quadrilaterals 321 5. 45 + x and 45 ■+- *. (Express the answer in terms of x.) 6. 60 + h and 60 -+- t-. (Express the answer in terms of u and V.) In Exercises 7-ll t the measures of the three angles of a triangle are ex- pressed in terms of a number %. In each case, find the value of .v and cheek your answer by adding the anjiie measures. 7. x, x — 5, x -s- 5 S. x, 2x, 3x 9. x + 1, x + 2, x + 3 10. x + 50, 2x + 30. 3x + 80 11, 3x + o»2r - 3, ox — 8 12. The figure shows a triangle A ABC, a point D between A and tf, the line CD, and the lines parallel to CD through A and B. What is the sum of the measures of angles 1 , 2, 5, and 6? How does the sum of the measures of Z 1 and Z 6 compare with the sum of the measures of £ 3 and Z4? How docs the sum of the measures of / 3 and Z 4 compare with the measure of £ACB? Use these ideas to write an alternate proof of The- orem 7.29, 13, The figure shows a parallelogram ABCD and one of its diagonals AC, D C What is the sum of the measures of angles LDAB and Z B? Why? What is the sum of the measures of angles LDCB and I D? What is the suni of the measures of the angles of a parallelogram? Why is AB ss CV, m m Z3A~ t and 55 SM A~C? Why is A ABC s ACDA? Use these ideas to write an alternate proof of Theorem 7.29. Prove Theorem 7,30, Prove Theorem 7.31. 16. Prove Theorem 7.32. (Start with two triangles, &ABC and AA'JS'C, and a correspondence ABC <s — > A'B'C) It. 15. 322 Parallelism Chapter 7 17. Justify the equations at the end of the proof of Theorem 7.33. 18. Prove Theorem 7.34. 19. Copy and supply reasons for steps 5, 7, 8 in the proof of Theorem 7.-35. 20. Let A ABC; and uurntan a, b. c be given. Suppose that m£A = ka, m£B = kh, m L C = kc for some number A% the same number k in all three cases, If a = 1,6 = 2,0 = 3, find mZA, mlB, mjLC. Check by addition. 21. Same as Exercise 20, except that a = 2, h = 3, c ss 5. 22. Same as Exercise 20, except that a = 100, h = 250, c = 300. 23. Same as Exercise 20, except that a s 180, h = 180, a = 180. 24. Same as Exercise 20, except that fits 1, h = 1, c = 100. 25. Same as Exercise 20, except that ft = 100, b = 1 , c = 100. ■ In Exercises 26-30. there is a labeled figure in which x denotes the measure of an angle and there is an assertion about .t. Justify the assertion. In the fig- ure for Exercise 30, y and z also denote angle measures. 20, i = 50 x = 90 27. ;v = 150 29. x = 115 30. x = 27 7,9 Thwrtms wi Triangles and Quadrilaterals 323 In Exercises 31-35, Ihcre is a labeled figure in which x denotes the length of a segment. At the left, there is an assertion about x. Justify the assertion. If ij appears in a figure, it also denotes a length. 31. x = 12 32. X = if 33. x = 10 34, X= 11 35. x = y 36. If the measure of the vertex angle of an isosceles triangle is 12.76, find the measure of each l>ase angle. 37. If the measure of each base angle of an isosceles triangle is 83, Dnd the measure of an exterior angle at the vertex of the triangle. 38. Find the sum of the measures of the exterior angles (one ul each vertex) of a triangle whose angles have measures as follows: 27,32* 59.39, and 93.2& 39. Given a parallelogram A BCD with AC — BD, find the measures of the four angles of the parallelogram. 40. Can a convex quadrilateral have three obtuse angles? Explain your answer. 324 Parallelism Chaptef ? 41. Given a convex quadrilateral ABCD with mZA = 90, find mLB + tnZC + mZD. Can LB and Z C both be obtuse angles? 42. Let AiJCD be aeon vex quadrilateral with m Z A = 90andmZB = 90. Can Z C and Z D both be obtuse angles? 43. L«t AS and CD intersect at E, an interior point of each segment. If AD = BC and AD || W, t prove that DC and AH bisect each other. 44. The measures of the angles of a triangle are 36, x t and y. We conclude that x is a number less than [7], tliat y is a number less than {T\, and that x + If = 13 ( In each case find the smallest number that will make the statement true.) 45. The measures af the angles of a convex quadrilateral are a, b> c, d. We conclude that each of these four numbers is less than [T|. (Find the smallest number that will make the statement true.) 46. The measures of the angles of a convex quadrilateral are 20, x, x, and y. We conclude that x is a number between [T| and [T|. that y is a number between £TJ and jT], (Make the smaller number as large as possible and the larger number as small as possible in each case.) 47. Given the plane figure in which AC is the bisector of Z BAD, CB X AB S and CD _L AD, prove that BC = DC. 48. Prove: If each angle of a convex quadrilateral is congruent to the angle opposite to it, then the quadrilateral is a parallelogram. [Hint: Given quadrilateral A BCD with LA^ LC % LBst Z£) } find»iZA + m/ H and deduee that two sides are parallel. Find »«ZA + mLD and de- duce that the other two sides are parallel.) 49. Given A ABC and A A DC with JD intersecting EC at M t an interior point of AD and EC as indicated in the figure, find the angle measures x, y, 3 u, t\ R«view Exercises 325 CHAPTER SUMMARY The central theme of this chapter is parallel lines, and the Iiigh point Of I he chapter is ihe PARALLEL POSTULATE. Before introducing Lhe Par- allel Postulate we proved several theorems on conditions which imply that lines are parallel These arc theorems in EUCLIDEAN GEOMETRY as wdl n in the NON-EUCLIDEAN GEOMETRY of Lobachcvsky and Bcdyai. These theorems are concerned with lines and TRANSVERSALS and angles associated with them. The Parallel Postulate gives us what we need to prove the converses of these theorems. PARALLELISM for lines in a given plane is an equivalence relation. that is, it is reflexive, symmetric, and transitive. The same is true for paral- lelism of segments in a given plane and for rays in a given plane. With parallel lines it is natural to associate PARALLELOGRAMS. This chapter includes several theorems regarding properties of parallelo- grams and several others regarding conditions tinder which quadrilaterals are parallelograms. An important consequence of the Parallel Postulate is the theorem on the sum of the measures of the angles of a triangle. An easy corollary of this theorem is the theorem on the sum of the measures of the angles of a convex qnadrilateraL Another important consequence of the Parallel Postulate is the result that parallel lines are everywhere equidistant Along with the idea of PARALLEL RAYS we introduced the idea of A NT! PARALLEL RAYS, There were several theorems regarding the re- lationship of two angles with parallel or anti parallel sides. Early in the book we introduced the S.A.S., A.S.A., and $.$*& Postu- lates. In this chapter there were two results which extend these congruence ideas, namely the S.AA. THEOREM and the HYPOTENUSE-LEG THEOREM for right triangles. REVIEW EXERCISES Copy and complete the definitious in Exercises 1-10. I, line m is parallel to line n if and only if [TJ, SL Two lines are skew lines if and only if (TJ- 3, Two segments are parallel if and only if \T\. A. Two collinear rays arc parallel if and only if jT|. 5. Two noncollincar rays are parallel If and only if [fj, & Two collinear rays arc antiparallel if and only if JT], 7. Two noncoUinear rays are antiparallel if and only if '|T]. 8. If distinct lines m, n t t are coplanar, then Ms a tramversal of m and n if and only if [TJ, 326 Parallelism Chapter 7 9. A parallelogram is a convex quadrilateral whose [?j. 10. The distance between two distinct parallel lines is [?]. ■ Exercises 1 1-20 pertain to the geometrical figure suggested by Figure 7-48. It is given that lines A& % CD, and EF arc parallel and noncoplaaar. Also, AB = CD = EF. In each exercise tell whether llic given statement is true or false, and explain why it is true Off why it is false. Figure 7-45 £ 11. There is more than one plane contacting A and B. 12. There is more than one plane containing A, B> and C. 13. There is exactly one plane containing A, B> E, and R 14 There is exactly one plane containing A, £, C, and F. 15. Quadrilateral ABDC is a parallelogram. 16. Quadrilateral ABFE is a parallelogram. 17. Quadrilateral CDFE is a parallelogram. 18. AACE St ABDF 19. A ABC =s ADCfl 20. m / £4 C+ in lACE+m£ CEA + m L FBD +ml. BDF+ m Z DVB = wZ CAB + m Z A£D+ mZ BDC+ m Z DCA Exercises 21-30 pertain to the geometrical figure suggested by Figure 7-49. Line ( is a transversal of lines m and n, The measures of eight angles formed by m, n, and f. have been marked in the figure. In each exercise, tell whether tile statement is true or false. If it is true* state a theorem that justifies your answer. Figure 7-49 Rftvtaw Exercises 327 21, If * = 125, then in [\ n, 26. If m ]n, then w = 125. 22, If u = 65, then in || n. 27. If n = 60, then m is not parallel to n. 23, If w = 125, then m | n, 28. If .v = MO, then m is not parallel to n. 24, If m '| n, then 3 = 125. 29. If y =£ u, then m is not parallel to n. 25, If m | n» then u = 65. 30. x = 126 31. In your own words write a sentence that tells of a significant difference between the geometry of Euclid and the geometry of Lobachevsky and Bolyai. 32. Explain what we mean when we say that the equals relation is reflexive, synunetric, and transitive. 33. Explain what we mean when we say that the relation of parallelism for lines is reflexive, symmetric, and transitive. 34. Is the relation of parallelism for rays an equivalence relation? ;15. h the relation oi antipardllelism For rays on equiva'.ru.-..- rln.n m.- In Exercises 36-42, you are asked to prove a statement. You may use any- thing that we have had in our formal geometry structure up to this point (except the theorem you are asked to prove) in writing your proof. 36. Prove that opposite sides of a parallelogram are congruent. 37. Prove that if two opposite sides of a convex quadrilateral are congruent and parallel, then the quadrilateral is a parallelogram. 38. Prove that if opposite sides of a convex quadrilateral are congruent, then the quadrilateral is u parallelogram. 39. Prove that opposite angles of a paralletogram are congruent 40. Prove that two consecutive angles of a parallelogram are supplemen- tary. 41. Prove that the two acute angles of a right triangle arc complementary angles. 42. State and prove the theorem regarding the angle measure sum of a triangle. 43. In the geometry of lobachevsky and Bolyai is it tnie that two alternate interior angles formed by two parallel lines and a transversal are congruent? 44. In the geometry of Euclid is it true that two alternate interior angles formed by two parallel lines and a transversal are congruent? 45. In the geometry of Lohachevsky and Bolyai is it true that it' two alter- nate interior angles formed by two distinct lines and a transversal of them are congruent, then the lines are parallel? 46. In the geometry of Euclid is it true that if two alternate interior angles formed by two distinct coplanar lines and a transversal of them are con- gruent, then the lines are parallel? Chapter Fritz Henlc/Fhoto Researclien Perpendicularity and Parallelism in Space 8.1 INTRODUCTION This chapter is concerned mostly with figures that do not lie in one plane. We use the word figure in two ways: as a set of points (geo- metrical figure) and as a picture or drawing that represents a geo- metrical figure. The figures in this book and the figures that you draw as you study geometry are important. In communicating information regarding geometrical figures one drawing may be worth 473 words. When drawing a figure, the perspective view is the t>cst one from the standpoint of communicating how the figure looks, and that is the view used in the illustrations in this book. However, the oblique view is the easiest to draw; therefore instructions for drawing figures in oblique views are given on the following two pages. A plane may be suggested by a parallelogram as in the figure which illustrates two intersecting planes. Note that the plane sug- gested by a parallelogram includes more than a parallelogram and its interior. It includes all of the points in the plane of lire parallelogram, the points in the exterior of the parallelogram as well as those on the parallelogram and in its interior. PROCEDURES FOR DRAWING GEOMETRIC FORMS IN OBlfQUE. VIEWS Draw front plane, edge or edges of form, in norma* elevation view. Piojeti from Hie comers, al a 45* angle, facetting Lid'd k h'..-,, wllos= unit of measurement sl Diid m H of those in Iht' elevation view. The bteck dashes ore guide lines to b* crated later. z Draw vertical gu.de lm« for the center of tne pyramid. / / y y* 76 330 A 7 / £JZ7 / ines that aro Inside or bor.irtd an obfect are by convention drawn 35 dashes. 331 332 Parpandlcularity and Parallelism Jn Space Chapter 8 A horizontal plane is oriented like the top of the cube. A vertical plane is oriented like the right face of the cube. Note Lhat horizontal and vertical are not defined formally in our geometry. They arc descriptive terms that we use to describe figures. The main topics in this chapter are perpendicularity and parallel- ism for lines and planes, in the preceding chapters of this book we have given detailed proofs for most of the theorems that were stated. Proofs not included in the text were assigned as exercises. In this chapter, however, we sometimes state theorems without giving proofs and without assigning these proofs in the exercises. In such cases we may include an outline of the main steps or a short statement of the strategy or main idea of a possible proof. In all cases it is possible to write detailed proofs based on the postulates of our formal geometry. Our abbreviated presentation is appropriate for a first course in formal geometry. 8.2 A PERPENDICULARITY DEFINITION Definition 8.1 A line and a plane are perpendicular if the line intersects the plane and is perpendicular to ever)' line in the plane through the point of intersection. If a line / and a plane a are perpendicular, we say that I is perpen- dicular to a and that a is perpendicular to /. Notation, / 1 a, or a 1 /, means that a and / arc perpendicular; I J_ a at P, or a 1 2 at P, means that t X a and that P is their point of intersection. Figure 8-1 represents a line / and a plane a that are perpendicular at P. The figure shows four of the lines through P and in a. Actually, there are infinitely many such lines and they are all perpendicular to I at P. riffureS-i 9.2 A Perpendicularity Definition 333 EXERCISES 3.2 ■ In Exercises 1-9, draw the figure that is described by the following statements. 1, Two distinct horizontal planes, 2. Two distinct vertical planes. & A horizontal plane and a vertical line, 4, A vertical plane and a horizontal tine. 5. A vertical plane and a line perpendicular to it 6* Two distinct parallel planes and a line that intersects both of them, 7. Two distinct parallel lines and a plane that is parallel to both of the lines. 8. Two distinct intersecting planes, one vertical plane and one horizontal plane. 9. Two congruent triangles tying in distinct parallel planes and the seg- ments connecting corresponding vertices, 10, See the figure. If i _L m, and m lies in a. docs it follow that / J_ a? Rxplain 1L In the figure, A-B-A\ AB = A% aX' I a at B t B $6 C, and C £ a. Prove that AC = A'C A \ ! / A' 334 Perpendicularity and Parallelism in $p*ce Chapter 8 12, In the following figure, aA' j. a at P, A'-P-A, A? = A'P, and B-D-C. On the basis of Exercise 1 i f what can you say about the lengths AB and A% AC and A'C, AD and A'D? Now prove that AABD = AA'BD, AACD s= AA'CD, A ABC s AA'JSC. 13. In the figure in Exercise 12, AT ± PB, AP ± PC; A'-P-A; B-D-C; a is the plane containing the noncollinear points B, C, P; AB — A'B; AC = A'C. State the postulate which implies that D is in a, Prove that AD := A'D. 14. {An informal geometry exercise*) Let I and m be two distinct lines in the plane of a table top and intersecting at a point R Hold a yardstick so that it appears to be perpendicular to I and to in at P. Use another yardstick to represent a third line n in the plane of the table top and passing through P. Docs the first yardstick appear to be perpendicular to lino n? Try n in various positions and see if you can find a position such that the yardstick and line n are no longer perpendicular. Does it appear that the yardstick is perpendicular to lino n in all cases? Does it appear that the yardstick is perpendicular to the table? 8.3 A BASIC PERPENDICULARITY THEOREM Theorem 8.1 is introduced to help us prove a basic perpendicularity theorem, Theorem 8.2. Theorem 8.2 is suggested by our experiences with perpendicular lines and planes. See Exercise 14 of Exercises 8.2, THEOREM 8. / If A, A ', B, C, D are distinct points with B and C each equidistant from A and A' and with D on BC, then D is equidis- tant from A and A f . Given distinct points A, A\ B, Q A with AB = A'B, AC = A'C, and D on BC as suggested in Figure 8.2, we want to prove that AD s A'D. Then D-B-Q or B-D-C 7 or B-C-D. The proof may be completed by showing that A ABC ^ AA'BC* AABD =* AA'BD* AACD a AA'CD. and hence that AD = A'D* 8,3 A Basic Perpendicularity Theorem 335 qp»u THEOREM 8.2 If a line is perpendicular to each of two distinct intersecting lines at their point of intersection, then it is perpen- dicular to the plane that contains them. Proof: Let lines l r m T n be given with m, n in plane a and such that I _L m at A y I 1 natA, as suggested in Figure 8-3. Let p be any line distinct from TO and n that ties in a and passes through A. We want to prove that I 1 p. Let q he a line in a that i s not parallel to m, or to n, or to p. and that docs not pass through A. Then q intersects to, p* rt in three distinct points; call them M, P, N, respectively. Let Q awl (r be two points of I such that Q-A-Q' and QA = Q'A, The proof may be completed by showing that M is equidistant from Q and Q\ S is equidistant from Q and Q', P is equidistant from {) and Q'j AQPA a AQ^PA, LQAP^ LQ'AP, and I X p. Since p is an arbitrary line oilier than to and n in the plane a and pass- ing through the point A, we have I X a. Vigun W 336 Perpendicularity and Parallel ism in Space Chapter 8 EXERCISES 8.3 1. Let the distinct points A, B, C, A E be given as suggested in the figure. A, B r C. D are coplanar points with no three of them eollinear. Also ED I W and ED 1 DC, Prove that ED 1 M. .?* 2. Let the distinct points A, H> C t D be given as suggested in the figure A, /*, C are noncollinear points in a plane «, S3 I a, and JJ = #C, Prove that IDAC a ZDCA, 3. Let the distinct points A, S t C, £> be given as suggested in the figure, ATS 1 J57), and AB _L HC. A, B, C are noncollinear points is a, and I) 4—* is a point not in a. Prove that AD is not perpendicular to a. 8.3 A Basic Perpendicularity Theorem 337 4, For the situation in Exercise 3, prove that AD is not perpendicular to tlie plane determined hy B, C D. Is AjB perpendicular to the plane BCD? 5, Let four noncoplan&r points A> B, C, D be given such that AB = AC = AD, A3 i plane ACD, AC 1 plane BAD, A& X plane BAG. Draw an ap- propriate figure and prove that A BCD is an equilateral triangle. 6, A rectangular solid has the property that if two of its edges intersect, then they intersect at right angles. The following figure shows a rcctan- gular solid with the eight vertices labeled. Prove that EA is perpendicu- lar to the plane that contains B t C, D. 7, Assume the same situation as in Exercise 6. Since EA is perpendicular to the plane that contains the quadrilateral ABCD t wc say that edge EA is perpendicular to face A BCD. Name six other combinations of an edge and a face that are perpendicular. (There are 24 such combinations altogether.) 8, In the proof of Theorem 8.1, suppose that D-B-C. Which Congruence Postulate would you use to show that AABCs AA'BC? That AABD Sf AA'BD? 9, Explain how the proof of Theorem 8.1 for the case in which B-D-C differs from the proof for the case in which D-B-C, Draw an appropri- ate figure. 10. Draw a figure like Figure 8-3, except much larger. Draw the segments connecting Q and Q with M, P, N. In completing the proof of Theorem 8.2, which theorem, postulate, or definition plays a key role in proving that (ft) M is equidistant from Q and Q' and that N is equidistant from Q and Q'? (b) F is equidistant from Q and p? (c) &QPA s AQ'PA? (d) LQAF=z Zp'AP? (e) / 1 p? 338 Perpendicularity and ParaKtali&m in Space ChaplerS 8,4 OTHER PERPENDICULARITY THEOREMS THEORFM 8,3 If a line and a plane are perpendicular, then the plane contains every line perpendicular to the given line at the point of intersection of the given line and the given plane. Proof: {See Figure 8-4.) Let / I a at P. Let t x be any line such that /, ^/atP. Figured We want to prove that ^ lies in «, Let be the plane that contains / and ly (See Figure 8-3.) Then /? ^ a since contains / and g does not. The intersection of (i and a is a line; call it / 2 . Then 1% _L J. Why? Also h ± I Why? Then h = k* Why? Therefore /, lies in a. Since ^ is any arbitrary line perpendicular to / at F, die proof is complete. figure. S-5 8,4 Other Perpendicularity Theorems 339 THEOREM 8.4 Given a line and a point, there is a unique plane perpendicular to the line and containing the point Proof: Lei £ be a given line and P a given point. We divide the proof into four parts. 1. If P £ /, we must prove that there is at least one plane perpendicular to / at P. 2. If P £ J, we must prove that there is at most one plane perpendicular to I at R 3. If P £ /, we must prove that there is at least one plane perpendicular to I and passing through P. 4. If P £ I, wc must prove that there is at most one plane perpendicular to I and passing through P. Parts 1 and 3 are existence proofs; parts 2 and 4 are uniqueness Proof of 1; If P is a point on a line 2 as in Figure 8-6, let rt be a plane containing I and let in be the line in a that is perpendicular to I at P. Figure Wl I^t Q be a point not in a and let fi be the plane that contains Q and L Let n be the line in /J that is perpendicular to / at P. Let v ta the plane that contains m and n. Then I A. y at E Why? This proves that there is at least one plane perpendicular to 1 at P. 340 Ptrpandteularity and Parallelism In Spac* Chapters Proof of 2: If point P lies on a line I as in Figure 8-7, we shall prove that there is at most one plane perpendicular to / at P. Suppose, eon- Figure S-7 trary to what we assert, that there are two distinct planes a and /? each perpendicular to I at P. Then a and /? intersect. Why? Their intersection is a line. Why? Call this line m. Let f and s be the lines such that r is in a and rimat F, and * is In /? and s 1 m at P, Let y be the plane that contains r and s. Then m J_ y at P (Why?), m X I at f (Why?), and I lies in y. Then % r, s are three distinct eoplanar lines with r _ I and s J. I, But this is impossible, There cannot be two distinct planes each perpendicular to I at P. Therefore there is at most one plane perpen- dicular to I at P, Proof of 3: Given a line / and a point P not on I as in Figure 8-8, let Q be the foot of the perpendicular from P to I Let a be the unique plane such that a ^_ I at Q. How do we know that there is one and only one such plane a? Then P lies in a. Why? Therefore there is at least one plane containing P and perpendicular to I . Figure B-9 8.4 Other Perpendicularity Theorems 341 Proof of 4: Given a line I and a point P not on it, as in Figures 8-9 and 8-10, we shall prove that there is at most one plane perpendicular to / and containing P. Suppose, contrary to our assertion, that there are two distinct planes a and ft each perpendicular to / and each con- taining P. Then the intersection of a and {S is a line; call it m. Figure S-9 Suppose I intersects m as in Figure 8-9. Then I intersects m in a point Q different from P. Why? Then a and /3 are each perpendicular to I at Q t But this is impossible. Why? Therefore I does not intersect m. Therefore there are two distinct points Q and K, as in Figure 8-10, such that I ± a at Q and I 1 £ at R. Then ARPQ has two right angles. Figure H-IQ But this is impossible. Why? Therefore there cannot be two distinct planes each perpendicular to I and each containing P. Hence there is at most one plane perpendicular to I and containing P. Definition 8.2 If A and B are distinct points, the unique plane that is perpendicular to AB at the midpoint of SB is called the perpendicular bisecting plane of AB. 342 Perpendicularity and Paralleflwt In Space Chapter 8 THEOREM 8.5 The perpendicular bisecting plane of a segment is the set of all points equidistant from the endpoinls of the segment. Let a be the perpendic- ular bisecting plane of A B and let C be the midpoint of A5. (See Figure 8-11.) Let P be a point. There are two things Lo prove, 1. IfPCa J thcnAP = ¥R %, If AP = PB,thenf€a, You are asked to prove these two statements as exercises. Figun* SI 1 THEOREM 8,6 Given two perpendicular lines, there is a unique line that is perpendicular to each of the given lines at their point of intersection. Proof: Let I and m be lines that are perpendicular to each other at P as in Figure 8-12. l^et a be the unique plane that is perpendicular to I at P, Let n be the unique line in a that is perpendicular to m at P. Then n is also perpendicular to L Why? Therefore there is at least one line perpendicular to both I and m at their point of intersection. Figure S-I2 Suppose that it and n' are distinct lines each perpendicular to m and to I at P as in Figure 8-13. Figure S-13 8,4 Other Perpendicularity Theorems 343 Let a be the unique plane that is perpendicular to I at P. Then n, n', m are three distinct lines in a and n 1 m, n' _L m. Since this is impos- sible, there is at most one line perpendicular to both I and m it then- point of intersection. Since there is at least one line and at most one line that is perpen- dicular to / and in at P t it follows that there is one and only one line that is perpendicular to both I and m at their point of intersection. THEOREM 8.7 If two lines are perpendicular to the same plane, they are parallel. Let I and m be lines that are perpendicular to plane n\ as in Figure 8-14- If I = m, then l\,m and there is nothing more to prove. Suppose, then, that I ^ m. Then it can be shown that I and m are non- intersecting lines. (If / and m intersect at a point A not in a, then A and the two points where I and m cut ct are the vertices of a triangle with two right angles. If I and m intersect at a point A in or, then the plane containing I and m intersects a in a line n such diat n ll,ni m,} Figure S-14 1 *-t F and Q be the points in which / and m, respectively, intersect a. Suppose, contrary to what we shall prove, that / and m are skew lines. I^t m* l>e the unique line through Q and parallel to I. Then / _L PQ and m* _ PQ. Why? Let t and & be die unique lines in a that are per- pendicular to r\) at F and at Q f respectively. Then I ± r and it fol- lows from Theorem 7.26 that m' _L ,v, Since m' J_ PQ and m* I 5, it follows that m' _L a. Let }i be the plane that contains m and »i' and let t be the line in which /? and a intersect. Then m _L £ m' J_ (, and the three distinct lines m f m', t are eoplanar. xSince this is impossible, it follows that I and m are not skew lines* Therefore they are coplanar nonintcrsecting lines, that is, they are parallel. 344 Perpendicularity and Parallelism in Space THEOREM S.8 Chapter 8 If one of two distinct parallel lines is perpendicu- lar to a plane, then the other line is also perpendicular to that plane. Proof: Given two distinct parallel lines / and m and a plane « per- pendicular to l t let P and Q be the points in which / and m intersect tx t as suggested in Figure 8-15, Let r s PQ, let s be the unique line in a perpendicular to r at Q, and let mf be the unique line perpendicular to r and s at Q (see Theorem 8,6). It follows from Theorem 8.2 that m' is perpendicular to or., from Theorem 8.7 that m* is parallel to h, and from the Parallel Postulate that m' = m. Since m f is perpendicular to a, it follows that m is perpendicular to a. •I Figure S-15 THEOREM 5,9 Given a plane and a point, there is a unique line containing the given point and perpendicular to the given plane. Proof: Let a plane a and a point P be given. We shall prove that there is a unique line through P and perpendicular to a. We divide the proof into four parts: two existence proofs 1 and 3 and two uniqueness proofs 2 and 4. 1. If P € a, we must prove that there is at least one line perpendicular to or at P. 2. If P £ <*. w c must prove that there is at most one line perpendicular to at at P. 3, If P g a, we must prove that there is at least one line perpendicular to ot and containing P. 4, If P g a, we must prove that there is at most one line perpendicular to a and containing P. 8,4 Other Perpendlculirity TTworems 346 Proof of 1: Given a plane a and a point P in a as suggested in Figure 8-16, let m and n be two lines in a that are perpendicular to each other at P. Then it follows from Theorem 8.6 that there is a unique line I that is perpendicular to both m and n at P. Then I -_ a at P. This proves that there is at least one line perpendicular to a at P. Figures- 1(1 Proof of 2; Given a plane a and a point P in a, as suggested in Figure 8-17, suppose contrary to what we shall prove, that m and n are dis- tinct lines and that each is perpendicular to a at P. Let /? be the plane that contains m and n and let I be the line in which /? intersects «. Then m l/,nl f, and the three lines I, m, n arc coplanar. This is impossible. Therefore there cannot he two distinct lines through P each perpen- dicular to a. Figure SI Proof of 3; Given a plane a and a point P not in a, as suggested in Figure 8-18, we shall prove that there is at least one line through P and perpendicular to a. Let Q be any point of a. If PQ J- a, there is noth- ing more to prove. Suppose, then, that FQ is not perpendicular to a. 346 Ptrjwndicularity and Parallelism In Sf»c« Chapter 8 Let I be the unique line such that IX a at Q. How do you know there is one and only one such line J? Let m he the unique line through P and parallel to /, Then it follows from Theorem 8.8 that m L <*. Therefore there is at least one line through P and perpendicular to a. Figure 848 Proof of 4: Let a plane a and a point P not in a he $ven. If there were two distinct lines through F each perpendicular to a, then there would be a triangle with two right angles. Since this is impossible, it follows that there is at most one line through P and perpendicular to a. EXERCISES S.4 1. Let four noncoplanar points A, B, C, D such that KB at "KC S6 AD, KB _L AC) AB _L A75, A7? JL ]S75 he given. Draw an appropriate figure and prove dial A BCD is not a right triangle. 2. I ,et four noncoplanar points A , B, C, D such tliat AB _L AT7, KB JL AD, 7VC _L AD be given. If A? is the bisector of L BAC, prove that / DAE is a right angle. 3. Let t m, n be three distinct lines, not necessarily coplanar, and such that 1 1| m and m |j n. Use Theorems 8.4, 8.7. and 8.8 to prove that I j| n. (This proves that parallelism for lines is a transitive relation.) 4. Let A, B, C t D, E, F be distinct points such that A-B-C, AB = BC, AD = DC, AE = EC, AF = FC. Draw an appropriate figure and ex- plain why B f E, D, F are copianar points. 5. Let segments AS and UD and three points E, W t G be given. If AT? and UD are perpendicular and intersect at E, if F-E-G and FT. L AB, docs it follow necessarily that FG is perpendicular to plane ABC? Draw an appropriate figure. 6. \jet segments AB and C/5 and three points £. F, C be given. If AB and CD are perpendicular and intersect at E, if F-E-G, TG* X AT?, and FD _ CD, does it follow necessarily that FG is perpendicular to plane A BC ? Draw an appropriate fiu,i IE& 8.5 Parallelism for Lines and Planes 347 7. In part 2 of the proof of Theorem 8.4, how do wc know that the inter- section of a and ft is a line? 8. In purl 2 of the proof of Theorem 8.4, how do we know that ml y at P? 9. In part 3 of the proof of Theorem 8.4, how do we know that P lies in a? 10. Given two distinct points A and B in a plane a describe the set of all points in a that arc eqtridistant from A and B. 11. Givon two distinct points A and B describe the set of all points that are equidistant from A and B. 12. Given a line / and the set § of all lines parallel to /, prove that if F is any point in space, then P lies on one and only one of the lines in $. 13. Give a plane a and the set § of all lines perpendicular to a, prove that if P is any point in space, then P lies on one and only one of the lines ing, 14. See the proof of Theorem 8.5. Prove that if P £ a y then AP = PB. 15. Sec the proof of Theorem 8,5. Prove that if AP = PB, then P £ a. IG. Given A ABC and two points DandE such that BA 1 AB, UA 1 AC, EE 1 AB : EB 1 EC, prove that A E, A, B are coplanar points, 17. See the proof of Theorem 8,9. In part 4 of this proof wc asserted that if there were two distinct lines through P each perpendicular to «, then there would be a triangle with two right angles. Draw an appropriate figure and prove this assertion. 8.5 PARALLELISM FOR LINES AND PLANES In this section we shall investigate the properties of planes parallel to planes and of lines parallel to planes. Wc begin with two definitions. Definition 8.3 Two planes are parallel if their intersection is not a line. Definition 8,4 A line and a plane are parallel if their inter- section is not a point. Let us consider Definition 8.3 first, If a and ft arc planes, then the intersection of a and ft is (1) the null set, or (2) a line, or (3) a plane. If* n ft = t then a^=ft and a \\ft. If a n ft is a plane, then « H ft = a = ft and a || ft. If a n ft is a line, then a is not parallel to ft. Therefore a is parallel to ft if and only if a = ft or a = . 348 Perpendicularity and Parallel*™ in Sface Chapter 8 It follows immediately from Definition 8,3 that parallelism for planes is reflexive and symmetric. Later we shall see that parallelism for planes is also transitive, and therefore an equivalence relation. The fact that parallelism tor lines is an equivalence relation follows from Definition 7.1, and Theorems 8.4, 8.7, and 8.8. See Exercises 10, 11, 12 of Exercises 7.6 and Exercise 3 of Exercises 8.4. Now let us consider Definition 8.4. Tf Ms a line and a a plane, then the intersection of I and of is (1) the null set, or (2) a point, or {3) a line. If I fl a = , then / 1 1 «. If / fi a is a line, then that line is t and 1 1 a. Ul n a is a point, then I and a are not parallel. Therefore I , a if and only if I C a or t H a = 0, THEOREM 8.10 If a plane intersects one of two distinct paral- lel hues but does not contain it, then it intersects the other line and does not contain it. Proof: (See Figure 8-19,) Let I and m be two distinct parallel lines. Let a be the plane that contains / and m, I^et /? be a plane that inter- sects one of the lines, say I, in a single point P. Now P lies in both ft and a. Since a contains / and $ does not contain /, it follows thai a =£ ft. Therefore the intersection of ft and a is a line; call it n. Since n lies in ft and I does not lie in ft t it follows that n is different from I. Also, m does not lie in ft. (Suppose m lies in ft. Then m lies in a and ft, and m = n. Since n is not parallel to l t it follows that m is not parallel to L Con- tradiction. Therefore m does not lie in ft,) Figure S-19 Therefore m, n, I are distinct lines in plane a with I parallel to m and with n intersecting I in a single point. It follows from the Parallel Postulate that n intersects m in a single point; call it Q. (Suppose n does not intersect m. Then n and £ are two distinct lines through P and parallel to m. Contradiction.) It follows that ft contains Q and hence that ft intersects m in a single point 8.5 Parallelism for Lints and Planes 349 THEOREM S. II If a plane is parallel to one of two parallel lines, it is parallel to the other also- Proof: Assigned as an exercise, THEOREM 8.12 If a plane intersects two distinct parallel planes, the intersections are two distinct parallel lines. Proof: Let a and ft be two distinct parallel planes and y a plane that intersects both of them, as suggested in Figure 8-20. Then y is distinct from a and from fi t and it intersects each of them in a line. Let / and m be the lines in which y intersects a and ft t respectively; then I and m arc distinct coplanar lines which do not intersect. Why do they not intersect? Therefore I and m are distinct parallel lines. Figure 8-20 THEOREM 8.13 If a, ft y are th ree distinct planes si ich that /? is parallel to y and such that a intersects /?, then a intersects y. Proof: (See Figure 8-2 L) Suppose that a, /?, y are distinct planes, that /? and y are parallel, and that a intersects 0, Figure 8-21 350 Perpendicularity and Parallelism In Space ChaptBf 8 Suppose, contrary to the assertion of the theorem, that a does not intersect y. Let m he the line in which a and /? intersect. Let P be a point of m; let I be a line through P that lies in a but not in ft and R a point of y. Let 8 be the plane containing I and R, Then 5 intersects $ in a line n distinct from m, and B intersects y in a line r. Then / and n are distinct coplanar lines through P and parallel to r. Since this con- tradicts the Parallel Postulate, it follows that a intersects y. THEOREM 8,14 If a line intersects one of two distinct parallel planes in a single point, then it intersects the other plane in a single point. Proof: Let a and ft be two distinct parallel planes and let I be a line that intersects a in a single point; call it F, (See Figure 8-22.) Let y be any plane that contains I, Then y and a arc distinct intersecting planes. Their intersection is a line; call it m. It follows from Theorem 8.13 and Theorem 8.12 that y intersects /? in a line; call it n. Then m and n arc distinct parallel lines that arc coplanar with I. Since / intersects m and is distinct from it, it follows from the Parallel Postulate that {intersects n and is distinct from it. Therefore the intersection of I and n is a single point; call it Q Since Q is a point of n, it is also a point of f$. Therefore Q is the unique point in which / and j8 intersect Figure 8-22 THEOREM H. 15 If a line is parallel to one of two distinct parallel planes, it is parallel to the other plane. Proof: Assigned as an exercise. THEOREM 8 JO There is a unique plane that contains a given point and is parallel to a given plane. 8.5 Parallelism for Line* and Plane* 351 Proof: (We give only a plan for a proof.) Let a point P and a plane or be given. Divide the proof into four major parts. 1. If P is in a, explain why there is at least one plane containing P and parallel to a. 2. If P is in a, explain why there is only one plane containing P and parallel to or. 3. If P is not in a, let / he the unique line through P and perpendicu- lar to a . Let ft be the unique plane that is perpendicular to I at P, Prove that a j /?, 4. If P is not in a, prove that there is at most one plane through P and parallel to a. THEOREM 8.17 Given a point and a plane, then every line con- taining the given point and parallel to the given plane lies in the plane containing the given point and parallel to the given plane. Proof: (Plan only.) Let a point P and a plane a be given. Suppose first that P lies ma. Then a itself is the unique plane containing P and par- allel to «. Show that the assertion of the theorem is true in tins case. Suppose nest that P is not in a. Let /? be the unique plane contain- ing P and parallel to «. Then a=£fi. I.et I be any line through P and parallel to a, as suggested in Figure 8-23. Use Theorem 8.14 to com- plete the proof. Fi'gure S-2rs EXERCISES SJ L If a and j$ are planes such that or n fi = 0, is n f /J? Why? 2. If / is a line and a is a plane and !n«= 0, is it possible that (a) t is parallel to a? (b) / is not parallel to a? Explain. 3* Let a; /?, y be three distinct planes with a parallel to # and a not par- allel to y- From which theorem of this section may we conclude that $ is not parallel to y? 352 Perpendicularity and Parallelism In 5p*c* Chapter 8 4. Prove Theorem SJLt. 5. Given a plane a and two distinct parallel lines I and m such that I is not parallel to a. From which theorem of this section may wc conclude that m is not parallel to a? 6. See the proof of Theorem 8.13. Draw a figure like Figure 8-21. Add some segments and a label to the figure to suggest the plane & 7. Prove Theorem 8.15. 8.6 PARALLELISM AND PERPENDICULARITY hi tins section we define measure of a dihedral angle, right dihedral angle, and perpendicular planes. This section includes, as you might have guessed, several theorems on parallelism and perpendicularity for lines and planes. Wc begin by repeating a definition from Chapter 4, Definition 4.20 If two noncoplanar halfpknes have the same edge, then the union of these halfpknes and the line which is their common edge is a dihedral angle. The union of this common edge and either one of these two halfpknes is a face of the dihedral angle. The common edge is the edge of the dihedral angle. Figure 8-24 suggests two dihedral angles, one of them appearing to be larger than the other. What does "larger" as used here mean? In the case of (plane) angles wc defined larger in terms of measure. Since we do not have a measure for a dihedral angle, we develop one. Figure S-24 Definition 8,5 The intersection of a dihedral angle and a plane perpendicular to its edge is a plane angle of the dihedral angle. 8.6 Parallelism and Perpendicularity 353 Figure $-25 suggests a dihedral angle A-BC-D, a plane a such that BC _i_ a, and L PQR, a plane angle of the dihedral angle A-BC-D. Figure s-as It seems plausible that all plane angles of a given dihedral angle should he congruent. This brings us to the next theorem. THEOREM 8. 18 Any two plane angles of a dihedral angle are congruent. Proof: Suppose that I ABC and LDEF are two plane angles of dihedral angle (2—Hl-J t as in Figure 8-26', We may suppose mat points have been picked in these plane angles and then labeled so that A and D lie in the plane GHI, C and F lie in plane IHJ, and AB = BC = DE = EF. It is easy to show that BCFE and BADE are parallelograms, that ACFD is a parallel- ogram, that AABCs A DBF, and hence IABC& LDEF, With Theorem 8.18 proved, it is now possible to define a measure for dihedral angles. Definition S. 6 The measure of a dihedral angle is ihe meas- ure of any one of its plane angles. Perpendicularity and Parallelism in Spact Chapter 8 Definition 8. 7 A right dihedral angle is a dihedral angle whose measure is 9G> Definition 8.8 Two planes are perpendicular if their union is the union of four right dihedral angles. THEOREM 8. 19 If a line is perpendicular to a plane, then any plane containing the given line is perpendicular to the given plane. Proof: Assigned as an exercise, THEOREM 8.20 If a line is perpendicular to one of two parallel planes, then it is perpendicular to the other plane also, Proof: Assigned as an exercise. THEOREM 8.21 If two planes are perpendicular, then any line in one of the planes and perpendicular to their line of intersection is perpendicular to the other plane. Proof: Assigned as an exercise. TfTEOREM 8,22 If two distinct intersecting planes are perpen- dicular to a third plane, then their line of intersection is perpen- dicular to the third plane. Proof: Figure 8-27 suggests two distinct intersecting planes a and fi perpendicular to plane y. Let I = AB be the line of intersection of a and fii let EC and BD he lines in which m and ft respectively, inter- sect v. Let h be the line in a that is perpendicular to BC at B. Figure 8-27 8.6 Parallelism and Perpendicularity Let ?2 be the line in /3 that is perpendicular to SD at B, It follows from Theorem 8,21 that h ± y and l 2 i y. The proof may be completed by showing that Ii ss-% a 1, and hence that / J. y< Definition &9 A segment, or ray, is perpendicular to a plane if the line which contains it is perpendicular to the plane, If a segment is perpendicular to a plane and otie end- point lies in the plane, then that segment is a perpendicular to the plane, and its endpoint in the plane is the foot of the perpendicular. Definition 8.10 If « is a plane and S is a set of points, then the projection of S on a is the set of all points (>, each of which is the foot of the perpendicular from some point of S, THEORFM 8.23 The projection of a line on a plane is either a line or a point. I-et t lie a line and a a plane. There are three cases to consider. 1. /lies in cr. 2. / is perpendicular to a. 3- / is not perpendicular to a and does not lie in <*. PrcH>fofCase 1: If J lies in a t then every point of / is its own projection on a t and therefore J is its own projection on a. Proof of Case 2: If / is perpendicular to a at the point Q, then the pro- jection of every point of t on a is the point Q, and therefore die projec- tion of / on a is the point Q\ Proof of Cam 3: If / is not perpendicular to or and docs not lie in a T let A, B, C be three distinct points of / that are not in a, and let A', B', C be their respective projections on a. It follows from Definition 8. 10 that AA' ± «, BB' J_ a, CC - a, and from Theorem 8J that AA' || BB\ BB' \ CC, AA' CC< Since / is not perpendicular to a and since A, B, C are distinct points, it follows that AA' r BB', CC* are distinct lines, T^t /3 be the plane that contains AA' and BB'. Then /3 contains the point C. Why? Since /? is parallel to S3' (Why?) and M' Is parallel to CC7, it follows from Theorem 8.11 that jS is parallel to £<?. Since fi contains C and is parallel to CC7, it follows that jS contains CC 1 . There- 356 Perpendicularity and Parallelism In Space Chapter S fore C lies in fi. Let f be the line in which /? and a intersect. Since A', B\ C all lie in a and in fi, it follows that they all lie on V. If / in- tersects a t say in point D, then D is its own projection on a, and since D lies in both a and /•?, it follows that D lies on V , Note that I is determined by A and B, and that V is determined by A' and B*. We have shown that if P is any point of I distinct from A and B (like C and D in the preceding paragraph), then its projection F on a lies on t. Conversely, if Q' is any point of l\ then Q* is the projection of some point Q on (, If I and a intersect in {X then Q? is its own projection and Q s £7. If Q' is a point of P not on L then there is a unique per- pendicular to a at Q' ? and this perpendicular intersects ( in the point Q which has the point Q* as its projection on or. It follows that every point of / has some point of V as its projection, and that every point of r is the projection of some point on I. Therefore the projection of I on « is the line V, COROLLARY 8.23.1 The projection of a segment on a plane is either a point or a segment. Proof: Let s be a segment, or a plane, and $' the projection of s on a. If s is perpendicular to «, then *' is a point If s lies in a r then s' = s and hence s' is a segment. Let I be any line not in a and not perpen- dicular to a. Let A, B, Che three points of I such that A-B-C and let A\ R, C be their respective projections on a. Then it follows from Theorem 8.23 that A' } B\ C are collinear and from Theorem 8,7 that A'-R'-C. (IfA'-B'-C were not true, then we would have distinct par- allel lines that intersect,) Since A-B-C on I implies A'-B'-C, we say that betweermess for points is preserved in the projection from ion a, If s = AC and s' = WU, then B is between A and C if and only if ff is between A' and C. Therefore s' is the projection of s. THEOREM 8.24 Given a plane a and two distinct points A and B su ch tha t AB is parallel to a t if AW is the projection of XB on a, then A'W^AB. Frtxtf: Assigned as an exercise, THEOREM 8.25 Given parallel planes a and /? and A A RC in a, if A', B', C are the projections of A, B, C, respectively, on j5, then AARC^&A'B'C. Proof: Assigned as an exercise. 8.6 Parallelism and Perpendicularity THEOREM 5.26 The shortest segment joining a given point not in a given plane to a point in the given plane is the perpendicular that joins the given point to its projection in the given plane. Roof: Assigned as an exercise. THEOREM 8.27 All segments that are perpendicular to each of two distinct parallel planes and have their endpoints in these planes have the same length. Proof: Assigned as an exercise. Definition 8.11 The distance between a point and a plane not containing it is the length of the perpendicular segment joining the given point to the given plane. Definition 8.12 The distance between two distinct parallel planes is the length of a segment that joins a point of one of the planes to a point of the other plane and is perpendicular to both of them. EXERCISES 8.6 1. There is no definition of berweenness for half planes in this book. Write your own definition for it. Using your definition, prove that if 3C lt 3C 2 , X3 arc half planes with 3C Z between 0Ci and 3Ca, then the measure of the dihedral angle formed by 5C X and 3C 3 and their common edge is the sum of the measures of the dihedral angles formed by 3C j and 3C2 and their edge and by 3Ca and 3Cs and their edge. 2. There are no definitions in this book for Ihc interior and the exterior of a dihedral angle. Write your own definitions for these terms. 3. Draw a picture of a cube and label its vertices. How many right di- hedral angles are there each containing two faces of the cube? 4. Given the labeled cube of Exercise 3, select one of the right dihedral angles and, using the vertices of the cube, identify two of its plane angles. 5. See the proof of Theorem 8.18- Draw a figure like Figure 8*26 and mod- ify it to show the quadrilateral ACFD. Prove that ACFD is a rectangle. 358 Perpendicularity and Parallelism In Spaca Chapter 8 B Exercises 6-16 are concerned with projections on a plane. In Exercises 6-14, sketch one or more figures and he prepared to defend your answer. 0, Is the projection of every triangle a triangle? 7. Is the projection of every ray a ray? H. Is the projection of every point a point? 9. Is the projection of every segment a segment? 10, Ls the projection of every line a line? 1 1. Is there an angle whose projection is a line? \ ray? A segment? A point? An angle? 12. ls there an acute angle whose projection is an obtuse angle? 13, Is there a right angle whose projection is an acute angle? An obtuse an- gle? A right angle? 14. Is there a segment that is shorter than its projection? linger? 15, If an edge of a cube is perpendicular to a plane, describe the projection of the cube on the plane. 1 5 raw a sketch, 10, challenge problem. If a diagonal of a cuIhj is perpendicular to a plane, draw a sketch of the projection of the cube on the plane. ■ In Exercises 17-23, draw an appropriate figure and prove the theorem, 17. Theorem 8,19 21. Theorem 8.25 18. Theorem 8.2G 22. Theorem 8-26 19. Theorem 8.21 S3. Theorem 8.27 20. Theorem S-24 24. In the proof of Theorem 8,22, show that h = fe = '■ 25. Let A-BC-D be a right dihedral angle with M 1 B€, EC 1 CD, AB = 3 v#, BC = 5, CD = 12, Prove that AD = 14. 26. The figure shows four noncoplanar points A, B, Q D such that AB _L AC, AB J. AT5. XC I AD, AB = AC = AD. Find the sum of the measures of I BDA, LBDC, L CDA. Chapter Summary 359 in Exercises 27-34, determine if the given statement is true or false. (An if-then statement is false if there are one or more instances in which the if-part is Lrue and the then-part false.) 27. If a line is perpendicular to two distinct intersecting lines at their point of intersection, then it is perpendicular to the plane that contains the intersecting lines. 28* If the intersection of a plane and a dihedral angle is an angle., then that angle is a plane angle of the dihedral angle. 29. If Z and n arc distinct intersecting lines and if I is parallel to plane « t then n is not parallel to plane a. 30. If T and n are distinct parallel lines intersecting (hut not contained in) two distinct parallel planes, then the planes cut off segments of equal length on the two lines. 3t< If two planes are perpendicular to the same line, they arc parallel. 32. If two Kltet are parallel to the same plane, they are parallel. 33* Given a plane a, the set of all points each of which is at a distance of 5 from a is the union of two planes each parallel to a. 34. If a is a plane and F is a point, there is one and only one plane contain- ing V ami parallel to a. CHAPTER SUMMARY We l>egan the chapter with some suggestions for drawing figures to represent geometrical figures that do not lie in a single plane. We used fig- ures throughout to help you visualize the relationships of lines and planes in space. Definitions of the following expressions were included. A LINK AND A PLANE .ARE PERPENDICULAR THE PERPENDICULAR BISECTING PLANE OF A SEGMENT PARALLEL PLANES A LINE AND A PLANE ARE PARALLEL A PLANE ANGLE OF A DIHEDRAL ANGLE THE MEASURE OF A DIHEDRAL ANGLE A RIGHT DIHEDRAL ANGLE PERPENDICULAR PLANES A SEGMENT (RAY) PERPENDICULAR TO A PLANE THE PROJECTION OF A SET ON A PLANE THE DISTANCE BETWEEN A POINT AND A PLANE THE DISTANCE BETWEEN TWO PARALLEL PLANES There are 27 theorems in this chapter. It is suggested that you write out these theorems and draw appropriate figures for them. 360 Perpendicularity and Parallelism In Space Chapter 8 REVIEW EXERCISES ■ In Exercises 1-15, copy and complete the given statement to obtain a the- orem of this chapter. 1. If a line is perpendicular to each of two distinct intersecting lines at their point of intersection, tl icn 3 2. Given a line and a point, there is a unique plane perpendicular [T]. 3. The perpendicular bisecting plane of a segment is the set of all points each oi' which is |TJ, 4. If two lines are perpendicular to the same plane, they are ]TJ, 5. If one of two distinct parallel lines is perpendicular to a plane, then \T\. 6. Given a plane and a point, there is a unique line containing the given point and |T|. 7. If a plane intersects one of two distinct parallel lines but does not con- tain ft, then it \TJ and does not contain it. 8. If a plane intersects each of Lwo distinct, parallel planes, the intersec- tions arc [7] lines. 9. If a, /?, y are three distinct planes such that a is parallel to /? and such that n intersects y t then [?]- 10. If a line intersects one of two distinct parallel planes in a single point, thenQ]. 11. If a line is parallel to one of two distinct parallel planes, then \T}. 12. Given a plane and a point, there is a unique plane that contains the given point and [?]. 13. Any two plane angles of a dihedral angle are [7}* 14. If a line is perpendicular to a plane, then every plane containing that line Is Q]. 15. If two distinct intersecting planes are perpendicular to a third plane, then their line of intersection is QJ ■ In Exercises 16-23, some lines or planes are described. In each exercise, state whether or not they must be parallel to each other, 16. Lines through a given point parallel to a given line. 17. Lines perpendicular to a given plane. IS. Lines perpendicular to a given line. 19. Lines parallel to a given plane. 20. Planes parallel to a given plane. 21. Planes perpendicular to a given plane. 22. Planes perpendicular to a given line. 23. Planes parallel to a given lino. Review Exercises 361 In Exercises 24-30, write a plan for a proof of the given statement. 24. If TF is the projection of A£ on a plane, then A'B' < AB. 25. If A'B' is the projection of AB on a plane a and if A'B' — AB, then AB | a. 26. If a and B are distinct parallel planes, if parallel lines m and n are in /?, and if lines m' and n f are the projections of m and n, respectively, on « T then m' is parallel to n'„ 27. If « and /J are distinct parallel planes, if $ and f are parallel and con- gruent segments in t and if s' and f arc the projections of * and t, re- spectively, on a, then $' and f are congruent. 28. If A$ I ifi?', AC 1 1 A 7 ?, and if A, B, C are noneollincar points, then plane ABC and plane A'B'C are parallel. 29. If a and /? are distinct parallel planes and if I and m are distinct parallel lines that intersect ft in points Li and Mi, respectively, and /? in points J-2 and I4& respectively, then LiAfiJUgLi is a parallelogram. 30. If every plane containing a given line is perpendicular to a given plane, then the given line is perpendicular to the given plane. er The Public Archives of Canada Area and the Pythagorean Theorem 9.1 INTRODUCTION You are familiar with the idea of area and have computed the areas enclosed by figures such as triangles, sqtiaras, and circles. As you know, areas are usually computed using numbers which arc lengths or dis- tances. You also know that a distance function is based on some seg- ment as the unit of distance. In informal geometry, we usually combine a number and a word in expressing a distance, for example, 4.5 ft. In formal geometry, we frerjuendy use the number by itself, omitting the unit when the distance function is understood. Tn our formal develop- ment of the area concept wc shall suppose that a distance function is given. Then there will be just one area function- In this chapter we adopt some postulates for area based on our ex- periences in computing areas and on our ideas about areas in informal geometry. These postulates include the formula for computing the area enclosed by a rectangle as well as statements of general properties that are useful in proving theorems about area. We also develop formulas for computing the areas enclosed by figures such as triangles and trape- zoids, and sve use area as a tool in proving the Pythagorean Theorem. 364 Area and the Pythagorean Theorem Chapter 9 9,2 AREA IDEAS Let us suppose that a unit segment for distance measure is given and that a square of side length I is given. We call this square the unit square. (See Fig- ure 9-1.) Just as distances and lengths are numbers associated with sets of points (for example, the length AH is associated with AH), areas are numbers associated with sets of points. The distance function matches a positive number with each segment. Simi- larly, the area function matches a positive number with each rectangle and also with many other simple figures, We shall agree that the area of the unit square is J , In practical ap- plications there is a system of areas for each system of distances. If the side length of the unit square is 1 ft., then the area of the unit square is 1 sq. ft. If the side length of the unit square is 1 cm,, then the area of the unit square is 1 sq, cm. (See Figure 9-2.) F%v«B-i 1 aq. Lin. Figure 9-2 1 so. in, When we speak of the area of the unit square as 1 , we are thinking of 1 as the measure of the set that includes afl of the points in the plane of the square that are inside of or on the square. (See Figure 9-3.) Strictly speaking, it would be better to speak of the area of die unit square region. We shall follow custom, however, and call it the area of the unit square. Similarly, we talk about areas of rec- tangles, triangles, and circles when we really mean the areas of the regions which are made up of these figures including their interiors. (See Figure 9-4.) When we say the "area of a triangle," for example, we mean the "area of the triangular region." Fljjure 9-3 9.2 Area Ideas 365 Definition 9.1 A polygonal region is a triangular region, or it is the union of a finite numher (two or more) of triangular regions such that the intersection of every two of them is the null set, or a vertex of each of them, or a side of each of them. Polygonal region Union of triangular run ions Figure 9-4 You may already know how to compute the area of a rec- tangle. If its sides are of length a and b, its area is ah, (See Fig- ure 9-5,) In Section 9.3 we adopt this formula for the area of a rectangle as a postulate. Let us see if it is really a reason- able and basic assumption to ure9-5 Figure 9-6 suggests our unit square and a rectangle with sides of lengths 3 and 4. The rectangular region is made up of 3 * 4 = 12 square regions each of the same Size as the unit square region. So the area of the rectangle should be 3 • 4, or 12. Similarly if a and b are any two natural numbers (that is, positive whole numbers), then the area of a rectangle which is a by b should be ab. □ Figure 8-6 I 366 Area and the Pythagorean Theorem Chapter 9 Figure 9-7 suggests our unit square and another square with sides of length ^, Let x denote the area of the square with sides of length L Since it lakes 9 squares of this size (3 rows with 3 squares in each row as suggested in the figure) to fill up the unit square, you can see that the area of the unit square should be 9 times the area x of the ^ by y square. Therefore 9x= 1 □ * — 1 x -$. l Figure 17 It appears, then, that the area formula, S = ah, is a reasonable formula for our ^ by j- square. Taking as| and h = -k we have 1 the area of a ^ by ^ square. Similarly, if n is any natural number, then area of a — by — square is — • — , or -=r. n Figure 9-8 shows a rectan- gle which is ^ by |. It is easy to see from die figure that the area of the rectangle should be 2 • 5, or 1 0, times the area of a } by 4- square; hence Therefore the formula tS as ah is appropriate for a rectangle which is | by |, Similarly, if a and b arc any two positive rational number*, then the area S of a rectangle which is a by b should be given by the formula S = ah. For example, if -_ i Ekpre 9-8 then 6 = |, I, _ 3 _ 24 " — 5 ~ 4I>* and if R is a rectangle which is a by £>, its area S is the sum of the areas 75 • 24 = 1800 Utde squares, each of which is 4 ' by -£$, Hence Since l|go = |=Jf 4=(lM). we see that the formula S = ah holds, in this case. 9,2 Area Ideas Let R be a rectangle which is a by b and suppose that either a or b t perhaps both, is an irrational number. Then, using the idea that an irrational number can be squeezed between two rational numbers that arc as close together as you desire, we can make it seem reasonable that the area S of R is again given by the formula S = ah, Figure 9-9 shows a rectangle R which is 1 by yf%, (Recall that \/2 is not rational and that i/2 can be represented by the nonrepeating decimal 1.41421 . . . .where the three dots indicate an infinite sequence of digits.) Thus L4< \/2<1.5> where 1 .4 and 1.5 are rational numbers, Now a rectangle 1 by 1.4 has area 1.4, a rectangle 1 by L5 has area 1.5, and therefore the area S of the 1 by \*2 rectangle R is somewhere l>ctween 1.4 and 1 .5, that \s t 1.4 < S < 1.5. 367 Figure 9-9 Continuing in this way* we can show that S is between 1.41 and 1.42, between 1.414 and 1.415, and so on. If you consider the sequence of areas of rectangles 1.4, 1.41, 1.414, . . . , you see that the sequence is increasing and the rational numbers in the sequence are getting closer and closer to \/2 through numbers that are less than \/2, Similarly, if you consider the sequence of areas of rectangles 1.5, 1.42, 1*415, . * . , you see that the sequence is decreasing and the rational numbers in the sequence are getting closer and closer to y/2, through muni tens that are greater than y/2. It seems reasonable, then, to con- clude that the area S of the 1 by \/2 rectangle R is given by Ssl'yfs y^. The foregoing discussion may be summarized as follows. If the sides of a rectangle are rational numbers, the rectangular region can be either subdivided into a finite number of unit square regions and the area obtained by counting, or subdivided into a finite number of square regions each — by — for some integer n > 2 and then counting and 368 Area and the Pythagorean Theorem Chapter 9 multiplying byi In both cases the result S is the same as that obtained by using the formula 5 = ah, where a and h are the lengths of two adjacent sides of die rectangle. If the side lengths are not rational, then we cannot subdivide the rectangular region into square regions with a rational side length and find the area by counting, or by counting and multiplying by -^- for some natural number a. We can, however, n 2 approximate the side lengths using rational numbers and consider the limit of the sequence of areas of the rectangles whose sides have lengths that are rational numbers. It can be proved that this limit exists and is equal to the product ah* In this book we adopt the formula S=al> for the area of a rectangle as a postulate. The ideas used to develop the formula S" = ah for the area of a rectangle are, of course, the same ideas with which you are familiar from your past experiences in informal geometry. They include the following: (1) every rectangle has an area, (2) congruent rectangles have the same area> and (3) area is additive in die sense that the area of a figure is the sum of the areas of the parts of the figure. Similar ideas may be used in developing area formulas for triangles, parallelo- grams, trapezoids, and so on. In Section 9.3 we state carefully our basic ideas about area- We call these statements die Area Postulates and use them to develop several area formulas. 9.3 AREA POSTULATES Our formal development of area in this chapter is limited to areas of polygons and is based on the following postulates. POSTULATE 27 (Area Existence Postulate) Every polygon has an area and that area is a (unique) positive nuxnlier. POSTULATE 81 (Rectangle Area Postulate) If two adjacent sides of a rectangle are of lengths a and b t then the area S of the rectan- gle is given by the formula S = ah. POSTULATE 29 (Area Congruence Postulate) Congruent poly- gons have equal areas, POSTULATE 30 (Area Addition Postulate) If a polygonal re- gion is partitioned into a finite number of polygonal subregions by a finite number of segments (called boundary segments) such that no two subregions have points in common except for points on the bound- ary segments, then the area of the region is the sum of the areas of the 9.3 Ares Postulates 169 Notation, If 3 is a figure, we frequently use |ff | to denote the area of ?. For example, &ABC\ denotes the area of A ABC. The area of quad- rilateral A BCD may he denoted by | quadrilateral ABCD\ or by \ABCD\ if it is clear that ABCD is a quadrilateral. Example J Figure 9-10 suggests a polygonal region ABCDEFG, Let us call this region (P. Segments CO and CP partition, or separate, (P into three subregions (Pi, (P2, (P3, where (Pi, is the union of quadrilat- eral ABCC and its interior, <S>z is the union of A CFG and its interior, and (P3 is the union of quadrilateral CDEF and its interior. According to Postulate 30 } the Area Addition Postulate, the area of (P is the sum of the areas of 6*1, (P & and (P3, that is, |<P| = ((P^ + |<Pa| + |(P 3 |. H C Example 2 Figure 9- 1 1 suggests a figure consisting of a square A BCD with its diagonals, AC and DB t and an adjoining isosceles right triangle, ABFC t with legs of length I, Let S denote the area of &BFC, S' the area of square ABCD, and S" the area of pentagon ABFCD. FindS, S\ S" using the Area Postulates. Figure U-ll Solution: It is easy to show that ABFC = A BEC SO that the triangles have equal areas and thai BFCE is a rectangle, actually a unit square. Then 1 = \BFCE\ = \&BFC\ + \ABEC\ = J auk; + \ABFC\ by the Rectangle Area Postulate by the Area Addition Postulate by the Area Congruence Postulate 370 Area and the Pythagorean Theorem Chapter 9 Therefore 2S = 1 and S = j. Since square ABCD is partitioned by its diagonals into four triangles, each congruent to A BFC, it follows from the Area Addition and Congruence Postulates that S' = 4S and hence that S' = 4 • j = 2. From the Area Addition Postulate it follows that S" = S' + S = 2J. Another way to find S' is as follows. Since ABbXl = ABEC, we have BE = EC=l and mlBEC = 9Q. Then it follows from the Pythagorean Theorem, which is proved in Section 9.5, that (BC)2 = 12 + 12 = 2 and BC = \M. Finally, it follows from the rectangle area formula that S' = y/2 • \/£ = 2- EXERCISES U ■ In Exercises 1-8, given the lengtta a and 6 of two adjacent sides of a rectan- gle, find the area of the rectangle. L a = 2,5, b = 3.3 5« = £ b = 4 2.a = % t b = i 0, a = 7, b = 3yl" 3.as|^s| 7, a = 4t/2, 6 = syi 4 a = 2,32, 6 = 1.77 8, a= Vo y b = 5^ 9. (An informal geometry exercise.) The length of a side of a square is 10 yd. What is the area of the square in square yards? In square feet? 10, (An informal geometry exercise.) ]f the dimensions of a rectangular field are 40 rods by 80 rods* how many acres of land does it contain? (An acre contains 43,360 sq. ft. and 1 rod is 16| ft. long.) 1L (An informal geometry exervim.) How many feet of fence would it take to enclose the field in Exercise 10? 12. (An informal geometry exercise.) The dimensions of a football field (including the end zones) are 53} yd, wide by 120 yd long. Would an acre of land be sufficient on which to lay out a football field? (See Ex- ercise 10,) 13. If the perimeter of a square is 4H, what is its area? 14* If the perimeter of a square is 64 \/2. what is its area? 9,3 Area Postulates 371 15- The length of one side of a rectangle is 3 1 hues the length of an adjacent side. If the perimeter of the rectangle is 5fi, what is its area? 16. The length of one side of u rectangle is 5 more than twiee the length of an adjacent side. If the perimeter of the rectangle is 70, what is its area? 17. challenck pkoblem. The length of a rectangle is 4 more than twice its width. If the area of the rectangle is 48, what is its perimeter? IS. The width of a rectangle is | of its length. If the area of the rectangle is 360, find its length and width. U). (An informal geometry exercise.) How many squares 1 in. on a side are contained in a square 1 yd. on a side? 20. (An informal geometry exercise.) How many 9 in. X 9 in. tiles would it take to cover a 12 \ ft. X 9 ft. rectangular floor if wc assume that there is no waste in cutting the tile? 21. If S is the area of an x by x square and S' is the area of a 2.v by 2.v square, prove that S" = 43, 22. If S is the area of a right triangle with legs of lengths a and b t and 5' is the area of an a by b rectangle, prove that S* = 2S. 23. If S is the area of a right triangle with legs of lengths a and b t and 5' is the area of a right triangle with legs of lengths 2a and 2b, prove S' = 4S. 24. The figure shows two coplanar equi- lateral triangles, AABC and AACP, with AC = 1 . The feet of the perpen- diculars from A and D to EC are E and F, respectively. Prove that ABCD is a parallelogram and that AEFD is a rectangle. 25. Complete the proof of the following theorem. THEOREM The area of a right triangle is one-half the product of the lengths of its legs. restatkmkst: Given: A right triangle A ABC with the right angle at C, AC=b 7 and DC = a. _ i Pram. |AAJ9C| = Proof: Let C be the point of intersection of the line through A and parallel to OB and the line through B and parallel to XU, (See the figure,) Why must these two lines intersect? YVhy is quadrilateral ACBC a rectangle? Complete the proof by showing that AABC ^ ABAC and that | A ABC = \ab. 372 Ar«i And the Pythagorean Theorem Chapter 9 9.4 AREA FORMULAS Figure 9-12 shows a parallelogram ABCD with points £ and F the feet of the perpendiculars from D and C, respectively, to line AB. Figure 9-12 The following definition has two parts. In (1) we define base and altitude, thought of as segments. In (2) wc define base arid altitude. thought of as numbers (lengths of segments, or distances). (Compare the corresponding definitions for triangles on page 265.) Definition 9.2 1 . Any side of a parallelogram is a base of that parallelogram. Given a base of a parallelogram, a segment perpendicular to that base, with one endpoint on the line containing the base and the other endpoint on the line containing the opposite side, is an altitude of the parallelogram corre- sponding to that base, 2, The length of any side of a parallelogram is a base of that parallelogram. The distance between the parallel lines con- taining that side and the side that is opposite to it is the cor- responding altitude. Of height In Figure 9-12 we call h the altitude, or height, corresponding to the base AB. The number h is the distance between the parallel lines A3 and DC. Similarly, the distance between the parallel lines AD and BC is the altitude which corresponds to the base AD, Let S denote the area of parallelogram ABCD in Figure 9-12, let AB = ft, and let CF s ft, Wc proceed to prove that the area S of the parallelogram is bh. THEOREM 9.1 If h is a base of a parallelogram and if h is the corresponding height, then the area S of the parallelogram is given by the formula $ = bh. 9.4 Area Formulas 373 talement L £DAE=& /CBF 2. lAEDj= IBFC 3. Z5E ss CF 4. AAED m ABFC 5. \AAED\ = \ABF€\ a \AFCD = S + |ABrcj 7. \AFCD\ = S+ |AA£D' 8. S= |AFCD| - |AAED| 9. |£FC£>| + AAED\ = \AFCD\ in. OC = b 1L |EFCD| = fc/i 12. fefi + |AAFD| = \AFCD\ 13. /?/i = |AFCD| - \AAED\ 14. S = Mi Reason 1. Why? 2. Why? 3. Why? 4. S.A.A, Theorem 5. Why? 8. Why? 7. Why? 8. Addition Property of Equal- ity 9. Why? 10. Why? 11. Why? 12. Steps 9, 11; substitution 13. Why? 14 Steps 8, 13; substitution In the remainder of this section we prove two more theorems re- area formulas. THEOREM 9.2 If b is a base of a triangle and if h is the corre- sponding altitude, then die area S of the triangle is given by the formula S = ^bh. Proof: (See Figure 9- 13 . ) Let A A3 C be given . Let 55 be the ray with endpoint B and parallel to AC. Let CE be the ray with endpoint C and parallel to AB. Let F be the point of intersection of bB and CE. Then ABFC is a parallelogram and A ABC SB A FOB. Figure 9-13 374 Area and the Pythagorean Theorem Chapter 9 Now b m A3 and h is the distance from C to A$> and so /i is also the distance between CE and S3, It follows from Theorem 9.1 that ABFC\ = bh> But I A ABC] = \AFCB\ (Why?) and |AABC| + |AFCB| = \ABFC\ (Why?). Therefore S + S = 6Ji, 2S = &/», and S = Jbfc. THEOREM 9.3 If 6i and b 2 are the lengths of the parallel sides of a trapezoid and if h is the distance between the lines that con- tain these parallel sides, then the area S of the trapezoid is given by die formula S s 4(&i + b*)h. Proof: Let trapezoid ABCD t with parallel sides AH and CD, be given as in Figure 9-14. h / A Figure 9-14 ______ — . , The lengths of these sides, or bases, are b\ and h% % and the distance between AB and CD is h. The distance to is called the height. Let E be the point on opp BA such that BE = Ify. Let F be the point on opp CD such that CF m b%. Then AEFD is a parallelogram (Why?) and its area is (bj + & 2 )n. We shall show that A BCD «— * FCBF is a congruence, Then it will follow that |ABCD| + \FCBE\ = \AEFD\ S + S = (hi + &_)fc S = •£(&! 4- &)fc. 9,4 Area For mulit 375 To see that ABCD «— * FCBE is a congruence, note that (1) LBAB = lEFC (5) XB E PC (2) ZABC a ZFCB (6) B'C <= GB (3) IBCD=* ICBE (7)C35 = B£ (4) LCDA = ABEF (8) DA sj EF This completes the proof. There are area formulas for other special polygons, but we shall not develop them here. In Section 9.5 we use area formulas us tools in prov- ing the Pythagorean Theorem. EXERCISES 9.4 I* If the area of a triangle is 40 and a base is S.. find the corresponding altitude, 2. If an altitude of a triangle is 12 and the area of the triangle is 62, find the base corresponding to the given altitude. 3. The lengths of the legs ol a right triangle are 17 and 9. Find the area of the triangle. 4* Find the area of an isosceles right triangle if the length of one of its legs is 15, 5. Find the area of a rhombus if its height is 7j and the length of one of its sides is 15^. ■ Exercises 6-8 are informal geometry exercises. 6. In the figure below, lines m and n are parallel and the distance between them is 5 ft. P and Q are points on m, and R and S are points on n such that FQ = RS = 7 ft. If the distance from P to R k 10 miles, find PQRS. P Q 4 — • ♦ 7/ btn 8 -M 7. In the situation of Exercise 6, find FQHS if PR = 1 mile. & In the situation of Exerd.se 6, find \PQRS\ if PR = 10 ft, 9* If the area of a triangle is S and its altitude is h, express the base b in terms of S and h. 10. The area of a parallelogram is 50.4. Find its base if the altitude is 4.2. 376 Area and the Pythagorean Theorem Chapter 9 11. Find the area of a trapezoid if its height is 7 and the lengths of the par- allel bases are 9 and 15* 12. Use the formula for the area of a trapezoid and express h in terms of 5, hi, and b& 13. If the area of a trapezoid ( s 128 and the lengths of the parallel bases are 7 and 9, find its height, 14. The area of a trapezoid is 147 and the lengLh of one of i ts parallel bases is 18. If the height is 7. find line length of the other base. 15. ABCD is a parallelogram with AB = 25. If \ABCD\ = 375 and P is a point such that C-P-D, find | AAPB|. Ifl. An informal geometry exercise.) A plot of land in the shape of a trapezoid has bases of lengths 121 yd. and 242 yd. If the distance be- tween the bases is 300 yd., how many acres of land are contained in the plot? (One acre contains 43,560 sq. ft) 17. Two triangles have the same base and their areas differ by 42, If the altitude of the larger triangle is 6 more than the altitude of the smaller one, find the length of the common base. 18. Id the figure, ABCD is a trapezoid with parallel liases AB and CD. If AB m fej, CD = h z , and the altitude is h, draw ED and use the Area Postulates and the area formula for a triangle to prove that \ABCD\ = ^(fcx + /ia). 19. In the figure, AM is a median of AABC. Prove that lAAJMfj = |AACMj. Is AABM r- A ACM in every case? Iu any case? The area of a square is equal to the area of a parallelogram . If the base of the parallelogram is 32 and its height is 18, find the perimeter of the square. 9.4 Ar»a Formulas 377 In each of Exercises 21-29, find, on the basis of the given data, the area or areas indicated, 21, X3 I Be* BC = 8, 23, ABCD is a square, DE = 6. f AAfiCJ = [T]. E is the midpoint of HC The perimeter of ABCD is 48, A D IAACE =m. 22. D-G-F r DF = 6, DE a 6.5, GE = 5. ADEF] = {T\. 26. ABCD is a parallelogram, E is the midpoint of C~D r and |A£CDj =33. JAA£D] =[?] jAA£E| =0] |ASCE| = fl 23, AH = CD = 8, BC = DA = 6, DE b 5. AABCD\ = JTj. 27. A£CD is a square; K, F, G, // are midpoints of the sides; jABCD, = 196, \EFGH\ = [?]. ABCD is a parallelogram, EF ± M, £F =3, BC= 8, CD = 4. |AA/*CD| = [2]. 378 Area and the Pythagorean Theorem Chapter 9 2H. EFGH is a square of side length 6; ABCD is a square of sick length 12; E, F s C, It are interior points of ABCD. The area of the shaded region 29. B and £ arc collinear with A and F ( K and H are collinear with L and (». C and / are collinear with B and /C, D and J are col- linear with E and H, AFCL is a rectangle with area 46, BCDE b ;i iert,mgjB with area 8, BCDE and jfKHJ are congruent rcctangjes. ABCDEFGHIJKL\ = [J fl J AT J H 30. Using the Area Postulates and the theorem of Exercise 25 in Exercises 9.3, derive the tri angle area formula for any triangle. 31. (See Figure 9-12.) Let b\ = AD and let A, be the distance between AD and BC. Is £?i/»i = hh? Justify your answer. 32. Would any of the statements in the proof of Theorem 9,1 be different if point £ were between A and B? Draw a figure for this case. 33. See the proof of Theorem 9.2. Prove that Q? and BD are not parallel lines. 34 In the proof of Theorem 9.2, prove that A ABC = AFCB. 35. Recall that a parallelogram is a special trapezoid. Show that the formula for the area of a trapezoid simplifies to the formula for the area of a parallelogram if the trapezoid is a parallelogram. 36. A triangle might be thought of as a quadrilateral in which one of the bases has shrunk to a point. What number does j[hi + b%)h approach if h t and h are fixed and h% gets closer and closer to 0? 37. challenge problem. Prove that the area of the triangle determined by the midpoints of the sides of a given triangle is one-fourth the area of the given triangle, 9.5 Pythagorean Theorem 379 9.5 PYTHAGOREAN THEOREM In this section we state and prove the Pythagorean Theorem and its converse. Perhaps no other theorem In mathematics has inspired more mathematicians and nonmathematicians alike to find original proofs as has this one. There are over 370 known proofs of this the- orem of which over 255 employ the use of areas. The first proof of this theorem has been attributed by many historians to Pythagoras (582- 501 Ji.c), a Greek mathematician and philosopher, although the prac- tical application of the theorem was known many years before his time. It is not known which proof Pythagoras gave, but in the Exercises at the end of this section is an outline of a proof which appeared altoul 300 b.c. in the first of the thirteen books of Euclid's Elements. Perhaps the most unique proof of the Pythagorean Theorem was devised by a 16-year-old schoolgirl, Miss Ann Condi t, of South Bend, Indiana, in 1938. The proof devised by Miss Condit and an original proof devised in 1939 by Mr. Joseph Zelson, an 18-year-old junior in West Philadelphia, Pennsylvania High School, show that high school students are capable of original deductive reasoning. For a list of proofs of the Pythagorean Theorem, including the proofs by Miss Condit and Mr. Zelson, see The Pythagorean Proposition by Elisha Scott Loomis, published by the National Council of Teachers of Mathematics, 1968. THEOREM 9.4 (The Pythagorean Theorem) If a, b, c are the lengths of the sides of a right triangle A ABC, with c = AB t the length of the hypotenuse, then a 2 + &>=<& Let BC = a, CA = h. Let PQRS be a square of side length a + b as in Figure 9-15. The points T, (/, V, W are interior points of PQ, QR* R§, SP, respectively, such that FT = QU = RV = SW = a. T b Q 380 Area and the Pythagorean Theorem Chapter 9 Then TQ = UR = VS = W? = h, and ATQU. AURV, AVSW, A WFT are four congruent right triangles, each with area ^ih. Hence TUVW is an equilateral quadrilateral. Since die two acute angles in a right triangle arc complementary angles, it follows that nuLWTU =180- nUPTW - nUUTQ - mlPTW-mlPWT - (mlPTW+nUFWT) z 180 - 90 = 90. By a similar argument it can be shown that the other three angles of TUVW toe right angles. Therefore TUVW is a square. Let x be its side length. Then (a + fr) 2 = \PQRS (a + fc) 2 = 4 AWPTj + \TUVW\ {a + bf = 4 * \ab + x 2 (a + &) 2 = 2al> + x 2 a 2 + 2d/> + fc 2 = 2ab + x 2 a 2 + fr 2 = x 2 . It follows from the S.A.S. Congmence Postulate that A WPT = AABG Therefore x = c and a 2 + fe* = e 2 . THEOREM 9.5 (Convene of The Pythagorean Theorem) If a, b, c are the lengths of the sides of a triangle and if a 2 + b 2 = c 2 , then the triangle is a right triangle and the right angle is opposite the side of length c. Proof: Let there be given AARC with AB - c, BC = o, CA = b, and a 2 + fc* = c; 2 . Let AA'B'C l>e a right triangle with B'C = a, CA' = b, A'B' = x, and the right angle at C" as shown in Figure 9-16. C o Figure SUA 9.5 Pythagorean Theorem 381 Then a2 + fez _ *2. Why? By hypothesis, a 2 4- h 2 = c 2 > By substitution, x 2 = c 2 and, since x > and c > 0, it follows that x-c. Therefore A ABC s AA'B'C'by the S.S.S. Postulate and LC^LC It follows that ZC is a right angle and that A ABC is a right triangle. When solving problems by use of the Pythagorean Theorem, it is often necessary to find the square mot of a number that is not a per- fect square. Recall that a rational number is a perfect square if it can be expressed as the square of a rational number. Thus 16 and 2 ,f are perfect squares because 16 = (4)2 and V = ($)*, whereas 32 and -^ arc not perfect squares. If we arc required to find the square root of a number like 32 or -^- t we can approximate the square root by means of a rational number or we can leave our answer in what is called simplest radical form, If % = y/A, where A is a posi- tive rational number which is not a perfect square, we say that x is in simplest radical form when it is expressed as B>/C t where B is a ra- tional number and C is a positive integer which contains no factor (other than 1) that is a perfect square. We make use of the following theorems from algebra when putting radicals in simplest form, 1. If a and b are positive numbers, then ■\/ab — Vff * \/5* 2. If a and b are positive numbers, then ft£_ Va V b yE' Example I Put (1) y/3£, (2) >/¥"' and (3) \/105 hi simplest radical form. i Solution: 1. V32 = \/W- 2 = y/TB • y/2 m 4 \/l" " / 8 " JT' 2 = X /I6 " 4 " 3. \/I06 is already in simplest radical form. Why: 382 Area and the Pythagorean Theortm Chapter 9 EXERCISES 9.5 In Exercises 1-15, put the indicated radical in simplest radical form. 1- V% 9, \/2M & \/!8 10. \/L35 3. v§7 4. y^O 5. v^S 6. yfffi 7. y300 11. vf 13. v*? x/ 12 ~rr 16. If ci 2 + b 2 = r; 2 , solve for a in terms of fc and c, 17. If o 2 + LP = c 2 , solve for b in terms of a andc. IS. If « 2 + b 2 = e 2 , solve for r; in terms of a and &, 19. If the length of a diagonal ttf a square is 15, find the area of the square. {Hint: Use the Pythagorean Theorem,} 20. In the figure, ABCD is a parallelogram, BE ± AB at E t AD - 15, CD = 20, and A£ - 9, Find the area olABCD. <> !■: 20 In Exercises 21-28, A ABC is a right triangle with right angle at C, a = BC, b = CA, c = AB, In each exercise, two of the three numbers a, b r c are given. Find the third one. Express each answer in decimal form correct to two decimal places. 21. a = 14 h = 1 jO 22- a = 1.0, b = 2.0 23. a= 10.0 f fo= 10.0 24. a = 3.0, 6 = 4.0 25. a = 9,0, b = 40.0 26. c = 3.1, fo = 5.2 27. a = 10.0, fc = 9.0 28. c ^ 10.0, a = 7,0 9.5 Pythagorean Theorem 383 In iLxereises 29-36, A ABC is a right triangle with right angle at C, a = BC t b = CA, G ss AM, In each exercise, two of the three numbers a, b, c are given. Find the third one, Kxpress each answer exactly and in simplest form, using a radical if needed. 29. a - 7, b = 40 30. a = 39, c = m 31. a = \/2, h = vT 32. = v/5I. fe = \ ^ 33. a = J, fc = i 34. o = 3x, £? = 4x (Find c in terms of *,) 35. a = x + gft & = x — y (Find c in terms of x and y.) 3ft. a = u 2 + C^ f a = a 3 — u 2 (Find b in terms of u and u.) 37. See the proof of Theorem 9.4, From which Triangle Congruence Pos- tulate docs the conclusion that ATQV s A URV follow? 38. See the proof of Theorem 9.4. From which Area Postulate does it fol- low that the area of the large square is the sum of the areas of the four triangles and the smaller square? 39. Sec the proof of Theorem 9.4. From which Area Postulate does it fol- low that the four triangles have equal areas? 40. In the proof of Theorem 9.4, which Area Postulate supports the con- clusion that the area of TUVW is x 2 ? 41. Let AABC with BC = a, AC == b, AB = v, c>b,c>a, and c 1 > « 2 4" b 2 be given. Answer the "Whys?" in the following proof that L C is an obtuse angle- Proof; Let AA'B'C He a right triangle with B'C = a, CA' = b, A'B' = x, and the right angle at C as shown in the figure. Then a* + t>i s s* for AA'B'C, Why? By hypothesis, a 2 + ^ < # for A ABC. By sub- stitution, x 2 < c 2 and, since x > and o > 0, it follows that x < c, Therefore IC> LC Why? Since mZC' = 90, it follows that m£C> 90. Theu L C is an obtuse angle. Why? 304 Area and the Pythagorean Theorem Chapter 9 42. Let A ABC Willi BC - a t AC = b, AB = c, c > b, c > a, and c* < a 2 + fe 2 be given- Prove that / G is an acute angle. (See Kxercise 41) 43. Combine Theorem 9.5 and the statement* proved in Exercises 41 and 42 into a single theorem. 44. chaijlenge problem. The Pythagorean Theorem may be stated as follows. In a right triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the legs, iiestatement! A ABC is a right triangle with the right angle atC, BC = a, AC = b, AB a c, squares ABDE, BCLK, and AIGC as shown in Figure 9- 17a. Prove Uiat t ABDE\ = \BCLK\ + \AFCX:\ and hence that c a = a 2 + h*. Complete the following proof which up» peared in Euclid's Elements around 300 b.cj. Figure ft.17 Proof: Draw AK, CD, and €37 1 IM at M and intersecting A~B at X as shown in Figure 9-1 7b. Why? (1) AABK ^ ADBC (2) \AABK\ = \ADBC\ (3) The altitude from A to £3 of AABK is equal to BC. (4) AABK = | * KB - »C = } • fl 2 Whv? (5) JBCLiq =2*| AABK | Wh^ff (6) The attitLide from C to B3 of ADBC is equal to BN. (7) 'ACBD, = l-PB-JJA' Whv? Why? Why? Chapter Summary 385 (8) BDMN\ = 2 * | ADBC\ Why? (9) \BCLK[ = \BDMN\ Why? (10) Draw BFand UE, ABAF = AEAC Why? (11) \AFGQ =2*|ABAf| Why? (12) |A£»fiV| = 2 • | A£Aq Why? (13) fAFCCf = [A£.\/,Y| Why? (14) ABDE = BDMN\ + \AEX1X Why? = \BCLK\ + AfGC| Why? (J 5) Therefore c 2 = « 2 + b 2 Why? 45- challexce problem. 11 your high school library contains the book 27*e Pythagorean Proposition, look up the proof devised by Miss Aim Gondii: and present it to your class. CHAPTER SUMMARY In this chapter there a re four Area Postulates and several area formulas stated ami proved as theorems. The last two theorems of the chapter are the Pythagorean Theorem and its converse. POSTULATES Area Existence Rectangle Area Area Congruence Area Addition FORMULAS Rectangle Area Formula S = ah Parallelogram Area Formula S s hh Triangle Area Formula S = \bh Trapezoid Area Formula S = -J(&i + b 2 )h THE PYTHAGOREAN THEOREM. If a, b, c are the lengths of the sides of a right triangle AA/JC, with c = AB> the length of the hypot- enuse > then a 2 + b 2 = c 8 . You should be able to stale each of the postulates in your own words and to explain the meaning of each of the formulas. You should be able to use the postulates in developing the properties of area. You should under- stand the formulas so that yon can recognize a situation in which a formula is applicable and then use it correctly. 386 Area and the Pythagorean Theorem Chapter 9 REVIEW EXERCISES 1. Find the area of a trapezoid with two parallel sides of lengths H and 13 if the distance between the lines containing those sides i- 24. 2. If the hypotenuse of a right triangle is 50 in. long and one of the lcg?i is 40 im long, how long is the other leg? 3. The figure shows a right triangle A.ABC with right angle at A and with D the foot of the perpendicular from A to EC. If BD - 3, DC = 5{, and Ai3 = 5, find \ A ABC 4. Find the area of a right triangle if the length of the hypotenuse is \/T and the length of one log is y/5. 5. Find the base of a parallelogram if its area is 143 and its height is 7. 6. Find the altitude of a triangle corresponding to a base of length 12 if the area of the triangle is 62, 7. If A ABC is an equilateral triangle of side length 6. find the area of the triangle. (f/iitfcUsc the Pythagorean Theorem to find an altitude of the triangle.) 8. See Exercise 7. Prove that the area S of an equilateral triangle of side length f is given fay ■ -J&.A 4 9. Use the result of Kxercise 8 to find the area of an equilateral triangle of side length 12, Compare the area of this triangle with that of Kxercise 7. Is the ratio of the areas of the two triangles the same as the ratio of the lengths of their sides? If not, how do the areas compare? 10. If ABCD is a parallelogram with m/ DAB — 45, DA a 12, and DC = 21, find \ASCD\. (Mint: Draw ~DE _ AB at E, Which kind of triangle is A DAE?) In Exercises 11-13, A ABC h given with a = EC, h = AC. and c = Afl In each exercise the three numbers a, b, c are given. Is L C a right, an ob- tuse, or an acute angle? (See Exercise 43 of Exercises 9.5.) 11. a = ft6 = g,cs 12 12. a = 8, b = 15, c = 17 13. a = ll,b= 13, c = 17 Ftevtow Exercises 387 14 *!Tie area of a trapezoid is 164 and the distance between the parallel bases is 10. If the length of one of the bases is 15, find the length of the other base. 15. (An informal geometry exercise.) The dimensions of a shower stall are 3 ft. by 3 ft. by 7 ft. How many 4 in. by 4 in. tiles are needed to cover throe of the four rectangular walls if wc assume that there is no waste in cutting the tile? 16. {An inform al geometry exercise, ) \ rectangt. Jar plot of land is 1 00 yd. by 150 yd. A standard city lot for this particular plot of land is 75 ft. by 150 ft. and sells for $3000- If a real estate agent can buy the plot of land for $30,000, how much profit would he make if he divided it into standard lots and sold all of thein? Disregard surveying and legal expenses, 17. An isosceles triangle has two sides of length 13. If the altitude to the base is 12, find the area of the triangle. 18. A rhombus has sides of length 10 and the measure of one of its angles is 45. Find its area. 19. The length of the hypotenuse of a light triangle is 25 and the length of one leg is 24, (a) Find the length of the other leg. (b) Find the area of the triangle. (c) Find the altitude to the hypoteuusc. 20. A right triangle has legs of lengths 10 and 24 and hypotenuse of length (a) Find the area of the triangle. (b) Find the altitude to the hypotenuse. 21. The length of one base of a trapezoid is .5 more than twice the length of the other base. If the area is 100 a net the altitude is 10, find the lengths of the two bases. 22. In the figure, AD is an altitude of A ABC. If AD = 24, AB a 26, and AC = 30, find BC and A ABC. 23. Recall that the diagonals of a rhombus are perpendicular to each other. Use this fact to find the area of a rhombus whose diagonals are of lengths 10 and 8, How is the area of the rhombus related to the lengths of its diagonals? State this relationship in the form of a theorem and prove it. 388 Area and the Pythagorean Theorem 24. In the figure, All _l W t B-C-D, AC = 10, BC FindHDand jAAG'D|, Chapter 9 AD = 17. 25. In the figure, with right angles and congruent segments as marked, if VA = 2andAB = 1, find VF. 26. Taking the area of a given unit square as I and using any of the Area Postulates except the Rectangle Area Postulate, prove that the area of a 4 by 5 rectangle is 20. 27. Taking the area of a given unit square as 1 and using any of the Area Postulates except the Rectangle Area Postulate, prove that the area of a i by i rectangle te £■ 28. The figure suggests an alternate proof of the Pythagorean Theorem which makes good use of the area formulas for a trapezoid and a tri- angle, Write out the main steps in this proof. (The proof suggested by this figure is attributed to President Garfield and was discovered by him around 1876.) Review Exercises 381 29. challenge problem. The figure shows a triangle and its three medi- ans intersecting at the point G which is two-thirds of the way from any vertex to the midpoint of the opposite side. Prove that 30. challenge problem. Given the same situation as in Exercise 29, prove that the six small triangles all have the same area. Courtesy uf Leo GatieJU Gallery Similarity 10,1 INTRODUCTION Our experiences with objects of the same size and shape suggest the concept of congruence in formal geometry. The idea of same shape suggests the concept of similarity which you are ftbouA to study in this chapter. Consider a picture and an enlargement of it. In the enlarge- ment each part has the same shape as it has in the original picture, but not the same size. The picture and its enlargement are examples oi similar figures. Consider a floor plan for a building. The floor plan is a drawing made up of segments labeled to show the lengths of the segments in the actual building which the segments on the drawing represent. The floor plan is similar to the actual floor of the building. Although the plan and the floor do not have the same size, they surely have the same shape. Each segment in this plan is much smaller than the segment it represents in the building. We express this fact by saving that the plan is a scale drawing of the floor. Each angle in the plan has the same size as the angle it represents on the actual floor. In drawing a floor plan, lengths arc reduced and angle measures are preserved, The lengths of segments in a floor plan are proportional to the lengths of the segments that they represent in the floor, and a state- ment of this fact is an example of a proportionality. In the next two sections we define a proportionality and develop some of its properties. 392 Similarity Chapter 10 In the remaining sections we develop the concept of similar figures. The main theorems of this chapter have to do with triangle similarity. The chapter includes a proof of the Pythagorean Theorem based on similar triangles. 10.2 PROPORTIONALITY Figure 10-1 shows two figures, one with side lengths labeled a, b, c, d, e and the other with a\ b\ c\ d\ e\ Note that a = 2a\ b = W t c = 2c\d = 2d", e = 2e\ and hence that JL — JL — .£ , d__ e_ __ p a' "" V " c' " d' " e f " dmH c-3 •- ?■ 9 c'*L5 d'mS *' = as a* 4 a'm2 e'«4 Figure 10-1 We can express the idea of these equations by saying that a, b, c, d r e are proportional to a\ b\ c\ d\ e\ respectively. Definition J 0.1 Let a one-to-one correspondence between the real numbers a, b, c, . . . and the real numbers a',h\ c f t ... in which a is matched with a', h is matched with b\ c is matched with c\ and so on, be given. Then the numbers a, h, c, ... are said to be proportional to the numbers a', b\ c', . . . if there is a nonzero number k such that a = ka\ b = kh\ c = k& The number k is called the constant of proportionality, in Definition 10.1, why do you think k is required to be a nonzero number? 10.2 Proportionality 393 Notation. We use = to mean "are proportional to." We use (a, b> c s . . .) to represent an ordered set of numbers. Then w means that the numbers a,b,c,... are proportional to the numbers d % h\d, t it being understood that a is matched with a\ h with f, c with c', and so on. Note that the order of the numbers in the set {a, h, c, , , ,) and in the set [a', b', c', . . .) is important only to the extent that corresponding numbers must appear in the same order in the pro- portionality. Thus if we have we could also write (h, c, a, . . .) = (bf, <f $ a\ . . .), and so on. Example 1 (See Figure 10-1.) (4, 7, 3, 6, 8) = (2, 3.5, 1.5, 3, 4). Example 2 (2, 3.5, 1.5, 3, 4) = (4, 7, 3, 6, 8). Since 2 = ^-4, 3.5 = 4-7, 1-5 = 4*3, 3 = ^-6, 4 = ^-8, the constant of proportionality in Example 2 is j. What is the constant of proportionality in Example 1? Example 3 If (*, y t 9) = (7, 8, 6), find x and y. Solution: There is a number k such that % = k ■ 7, y = k • 8, 9 = k • 6. From the last equation we have k = j, so 3 = 4-7 = 10.5 and y = §-8=12. Example 4 U the following statement true or false? (5, 8, 10) = (8, 12.8, .16). Solution; The statement is true if and only if there is a nonzero num- ber k such that 5=Jfc-8 8 = it- 12.8 10 m k* 16 From the first of these three equations, we see that k must l>c -|, or 0.625. We must now check to see if this k "works" in the other two equations, 394 Similarity Chapter 10 Check. Is 6 = 0,625 - 12.8? Yes, Is 10 = | • 16? Yes. Therefore (5, 8, 10) = (8. 12.8, 16) is true. Example 5 Ts the statement (4, 8, 10) = (6, 9, 16) true? Solution: If 4 = k • 6, then Jfc = f , Is 6 = j< 9? Yes, Is 10= | -16? No. Is the statement (4, 8, 10) = (6, 9, 16) true? No. EXERCISES 10.2 In Exercises 1-12, determine whether the given statement is true or false. If it is true, find the constant of proportionality. The answer for Exercise I has been given as a sample. 1. (1, 1, 5) = (3, 3, 15). TYue; k = \. E. (3, 6, 9) = (2, 4, 6) 3. (5, 1) = (8. 2) 4. (8, 10, 15, 21) = (7, 1 1. 16, 22) 5. (5, 15, 25, 35) = (1, 3, 5, 7) 6. (2, 6, 7, 15) = (1, 3, 3.5, 7.5) 7. (1,3.1 +3) = {2, 6, 2 + 6) 8. (x, y. % + y) = (7x, ly t 7x + 7y) 9. (m, v,w) = {-3«, -3t\ -3m>) 10, (0, 0, 0) = (5, 10, 15) U. (5, 10, 15) = (0, 0, 0) 12. (0, 0, 3) = {0, 0, 4) In Exercises 1 3-20, copy and complete the given statement so that it will be true. 13. (8, 8) = (0,6) 14- mm, 100) = (5, 6, 10) 15. (5, 0,7) = ([7], 8, 10.5) 16. If (3, x) = (8, 10), then x = [7J. 17. If (7. 9) = (1x, 18), then a = [7J. 18. If (7. x) = (14, 21), then x = \t\> 19. If (5 t i) = {*, 125) and * > 0, then x = [7]. 20. If (3, x) = (8, if) and i/ yt o t then {3, [3 ) = (*, y). 10.2 Proportionality 395 In Exercises 21 -30, x is a positive number and (3. x) = (5, 15). In each ex- ercise, determine whether the given statement is true. 21. £ = ^- 5 15 22. — = — x 15 23. (3,*.3 + *) = {5, 15*20) 24. {x, 3 - x) = (15. 10) 25. (*, 3) f (15, 5) 20. (3, 5) = (x, 15) 27. (3. 15) = fc 5) 28. (3 t 5) = {15\r) 29. 3-15 = x-5 30. 3 5 x + 3 5+15 31. Does (fl, &, c) * — » (the,/) indicate the same one-to-one corre- spondence as (a, c, b) * — * (d, /, *?)? As (c, a, fo) « — * (/, rf T e)? As fcfea) *— * (/.«.d)? 32. If a, ft, e, <i are nonzero numbers such that (a, h) = [c, d) with pro- portionality constant 2 , is it true that (b,a) = {d, c) with proportionality constant 2? Give your reasoning. 33. If a, b, c,d are nonzero numbers such that (a, b) = {c, d) with propor- tionality constant 2, is it true that (c, d) = (a f b) with proportionality constant 2? Give your reasoning. 34. If a t b, c, d are nonzero numbers such that (a, b) = (c, </), prove that (a, c) = {b, d). 35. If {a, b) = (c, if), prove that ad = he, 36. If ad = fee, prove that {a, b) = (c, d). In Exercises 37-40, copy and complete I he given statement so that it will be a proportionality. 37. (5.7,10.12) = (10, [Sm, 0) 38. (0, H, 1) f (15* 37, 100) 39. (0,37. 0.67, 0.93) = ((?], 67, fTJ) 40. (V5 v^V^f CV&EH3) 396 Similarity Chapter 10 10.3 PROPERTIES OF PROPORTIONALITIES As you might expect from your study of algebra and from some of Exercises 10,2, proportionalities have some interesting properties. In this section we show that the proportionality relation is reflexive, sym- metric, and transitive, and therefore it is an equivalence relation. Hence the relation denoted by **=" has some properties in common r with the relations denoted by " = " and "^." We shall also state and prove addition and multiplication properties. At the end of this sec- tion we consider some special proportionalites called proportions. We shall prove the next two theorems for proportionalities involv- ing triples of numbers. It is easy to see how the statements and proofs can be modified for proportionalities involving more than or fewer than three numbers. THEOREM 10.1 The proportionality relation is an equivalence relation. Proof: 1. The Reflexive Property. Let a, b* c be any real numbers. Ob- serve that (a, b f c) == (a, b, c) with proportionality constant 1. 2. The Symmetric Property. Suppose that {a, h % c) — (*£» e,/). Then diere is a nonzero number k such that a = kd, b = ke t c = kf. Why? Then d = k'a, e = k'b, f= k'c, where k' m l and k' ffe 0. Therefore (d, <?,/) = (a, b, c). Thus if (a, b, c) = (d t ej) with constant of proportionality k t then (d. e t f) = (a, b, c) with constant of proportionality — . 3. The Transitive Property. Suppose (a, b* c) = (d t e t /) and {d, e f f) = (g, h y i). Then there are nonzero numbers kj and k 2 such that a = kid, b = k\e, c = k\f, d = k 2 g, § = k 2 K f = M- Then a = kid = fci(feg) as (hk 2 )& b = k x e = ki(k 2 h) = (hk 2 )h, c= kif=ki(k 2 i) = (k t k 2 )i. 10.3 Properties of Proportionalities 397 Since In jfi and k z # 0, it follows that Mfc f£ 0, Therefore, if a, b t c are proportional to d, e, / with constant of proportional- ity &i, and d, e./are proportional to g, ft, i with constant of pro- portionality fca, then a, b, c arc proportional to g, h, i with con- stant of proportionality fci& 2 - In the next theorem note that (1) is an "addition" property and that (2) is a "multiplication"' property. THEORRV 10.2 If (a, b, c) =: (</, e f /), then (1) ( and, if h ^ 0, (1) (a, fe, c. a + b + c) = {d t e t f, d + e + f) y (2) {ha t b>c) = {hd>e t f). Proof: Let it be given that (a> b, c) = (ef t e, /). Then there is a non- zero number k such that a = fcd, b = fee, c = Jfcf. Adding, we get a + b + c e faf + he + fcf, and, by the Distributive Property, a + b -*- c = k(d + e + /). Multiplying both sides of a = kd by h, we get ha = h(M), and hence ha =s Jt(fcd). We have shown that if a. b % c are proportional to d, e, f with constant of proportionality k, then (1) and (2) hold with the same constant of proportionality. Example 1 Figure 10-2 shows a triangle A ABC and a segment DE joining a point D of AB to a point £ of AC. Suppose we know that {AD, DB) = (A£, EC). Figure 10-4 398 Similarity Chapter 10 We may conclude from the addition property of proportionality that {AD, OB, AD + DB) = (AS, EC, AE + EC) and hence that (AD, DB, AB) = (AE, EC, AC), since AB = AD + DB and AC = A£ -f- EC. Of course, we may also conclude that (AD, AB) j AE, AC). This may not be the "whole truth/' but it is certainly the "truth." It is like concluding that if x = 3», y — 3u, and z = Bm, then * = Bit and y = Sv. Similarly, we may conclude that (DB, AB) = (EC, AC), Example 2 Let A ABC with D an interior point of BC be given as suggested in Figure 10-3. Let /* denote the distance from A to BC, let BD = bu and let DC = b* Then | AABD\ = * 2 hbu \AADC\ - $hb 2 . Figure 104 B Therefore (\AABD\, \AADC\) = (bi, bz), and the constant of propor- tionality is -j/i. Thus the areas of the two triangles formed from A ABC by inserting the segment AD are proportional to the lengths of the seg- ments formed from EC by inserting the point D, In this connection some of you will recall Challenge Problem 29 in the Review Exercises of Chapter 9, We frequently work in geometry with proportionalities in which two numbers are proportional to two numbers. Such proportionalities are generally called proportions. Following a formal definition is a list of some of their special properties. Definition 10.2 If a t b, c, d are numbers such that (a> b) =: (c, <l) is a proportionality, then that proportionality is a proportion. 10.3 Properties of Proportionalities 399 lii other words, a proportion is a proportionality with two num- bers on each side of the " — " symbol. The following theorem includes four named properties of proportions. THEOREM 10.3 Proportions involving nonzero numbers a, h, c, d have the following properties: 1. Alternation Property. If (a, b) = (c, d), then (a, c) = (b t d) and (cf, h) = (c, a). 2. Inversion Property. If (a , fr) =3 (e» d) % then (h a) = (</, o). 3. Product Property, {a, b) = (c, d) if and only if ad = be. 4 Ratio Property, (a, b) = {c, d) if and only if -y = ■£ . p b a Proof: Assigned as exercises. If you think of a and d as the "outside" numbers and b and c as the "inside" numbers of a proportion (*. *) f te A the Alternation Property says that if you interchange either the out- side numbers or the inside numbers in a proportion, then the result is a proportion. If you combine the Inversion Property with the Ratio Property, the Inversion Property amounts to saying that if two ratios are equal, then their "inversions" (reciprocals) are equal. In Theorem 10.3, the numbers in the proportions arc required to he nonzero numbers. What would lie the situation if zeros were per- mitted? Note that (0, 5) = {0, 10} is a true statement, whereas (0, 0) = (5 T 10) is a false statement. Therefore the proportion (0, 5) = (0, 10) does not have the Alternation Property. Observe that « 7 s • 8 is a true statement, whereas (0, 0) = (8, 7)isa false statement. Therefore the Product Property does not apply to • 7 = ■ 8. Observe that a,0) = (2,0) is a true statement, whereas ^ = -^ is a false statement. Therefore the Ratio Property does not apply to (1, 0) = (2, 0). 400 Similarity Chapter 10 EXERCISES 10.3 1. If 0, b t c t d are all nonzero numbers, then (a t h t c t d) — (a, &* c, d). What is the constant of proportionality in this proportion? Which prop- erty of an equivalence relation does this illustrate? 2* Given {a, b, c, d) = faj\ g> h), prove that fa f, g. h) = fa b> c d). Which property of an equi valence relation does this illustrate? What is the relation between the constants of proportionality for these two proportionalities? 3. Given (a, b) = fa d) with proportionality constant k\ and (c, d) = faf) with proportionality constant k->, then fa b) = fa f) with some pro- portionality constant, sny k^. How are k±, hit, k-t related? Which prop-- erty of an equivalence relation do we have illustrated here? 4. In the figure A, B, C are noncollinear points; A, B\, B& B are eollinear and arranged in the order named; A, C u C 2i C are colUneur and ar- ranged in the order named. It is given that (AB U »,B 3 . B 2 B) = (AC,, C,C 2 , dQ with constant of proportionality 0.8. Prove, using the properties of proportionalities, that (a) (AB,,AB) = (AC lt AC). (b) (16AB U AB) = (I6AC1, AC), In Exercises 5-11, complete the given statement and name the property which it illustrates. Assume that none of the numbers in these exercises is zero. 5. If fa b) = (3, 5), then (a, 3) = (£>, 0). 6. lf(x,i/) = (6 3 7) t then ( ?/ ,,v) f (7,0). 7. If (m. c) = (5, 6) and (5, 6) = (x, tj), then [u, v) = (x, [?]). 8. If (a. b) = fa 00), then fa a + b) = fa c + d). 9. If (*, y) = (I, 2X then (3*. y) = {[[J, 2). 10. If (x, y) = (J7J, 7), then 7* = 4i/. 11. If (*, y) = (5, 8), then * = [B 10,4 Similarities between Polygons 401 In Exercises 12-21, write a proportion, or complete the given one, so that it will be equivalent to the given information. Starting with the given in- formation, you should be able to prove that the proportion is true. Starting with the proportion, you should be able to prove that the given equation (or equations) is true. 16. a = 3*. b = 3y 21. ^ = - Exercises 22-25 refer to Theorem 10,3. Prove that proportions involving nonzero numbers have the indicated properties. 2SL The Alternation Property. 23. The Inversion Property. 24. The Product Property. 25. The Ratio Property. 10.4 SIMILARITIES BETWEEN POLYGONS In this section we define what is meant by a similarity between two polygons. In developing our formal geometry we use similarities as tools; in most cases we consider similarities between triangles. Definition 10.3 1. A one-to-one correspondence ABC . , . * — > A'B'C , . . between the vertices of polygon ABC . . . and polygon A'B'C ... is a similarity between the polygons if and only if corresponding angles are congruent and lengths of corresponding sides are proportional. 2. If ABC . . . < — * A'B'C ... is a similarity, then polygon ABC . . . and polygon A'B'C , , . arc similar polygons and each is similar to the other, 3. If ABC . . . a — * A'B'C ... is a similarity with A3 = kA'B', BC = kB'C, and so on, then k is the con- stant of proportionality, or the proportionality constant, for diat similarity. 402 Similarity Chapter 10 Notation, The symbol "W is read "is similar to"; hence ABC ,..*-» A'B'C ... is read "ABC . , . is similar to A'B'C . ..." This means that A IH A'B'C Example 1 Fi give 10-4 shows two quadrilaterals ABCD and A 'B' CW with segment lengths and angle measures as indicated. It ap- pears that ABCD * — * A'B'CD' is a similarity and hence that ABCD ~ A'B'CD'. Let us check this conjecture. There arc eight pairs of corresponding parts including four pairs of corresponding angles and four pairs of corresponding sides: LA and I A'; Al5 and AT*'; B and L W-. BE and JfC; LC and AC; UD and CT?j LD and L D*; UA and /XV. D A C DO 1 I34X 90 1 X Ag/^ A 7 B P FIgun KM First we check the angles. mZ.A = niLA' = 90, mlB s mLB' = 45, mLC = mLC = 135, mLD = mLD' = 90. Therefore, corresponding angles are congruent. Next we check to see if the lengths of corresponding sides are proportional. We want (AB, BC, CD. DA) = (A'B f . B'C, CD\ &A% Substituting, we get (7,3x/2*4, 3) = (3.5, 1.5 y/2, 2, 1.5). Since this is indeed a proportionality, with constant of proportion- ality 2, the lengths of corresponding sides arc proportional. There- fore, it follows dircctJy from the definition of a similarity that ABCD- A' B'C D\ 10.4 Similarities between Polygons 403 In Example 1 the constant of proportionality between the lengths of the sides of quadrilateral ABCD and the lengths of the correspond- ing sides of quadrilateral A'B'CD' is 2. Is it possible that in another example the constant of proportionality might be 1? Of course it is. In this special case of a similarity, corresponding angles are congruent and corresponding sides are congruent. Hence for this special case you should sec that Lhe similarity is a congruence. In other words, a congruence between polygons is a similarity between polygons for which lhe constant of proportionality is L Just as congruence for triangles is an equivalence relation , so also is similarity for polygons. We slate this fact as our next theorem, THEOREM 10 A The relation of similarity between polygons is reflexive, symmetric, and transitive. Proof: We shall prove the theorem for triangles. Tt is easy to modify this proof to get a proof for quadrilaterals* pentagons, and so on. Reflexive. Let AABC be given. Since AABC at AABC, it follows that A ABC — A ABC. What is the constant of proportionality for (AB, AC, BQ = (AB, AC, BQ? Symmetric. Suppose that AABC — ADEF; then ZA^ZA ZB^ZJE, ZC==ZF,and {AB, BC, CA) = (DE, EF, Dl<)> We want lo prove that ADEF — AABC, which means that ZD^ LA, LE^i LB, LF= / C, and that (DE, EF, DF) = (AB, BC, CA). Now LA=*LT) implies LD^LA, LBsLE implies LM m£B, and £Cm£W implies LF = Z C (Which property of congruence for angles supports tills deduction?) Also, (AB, BC, CA) = (DE, EF, DF) implies Hi at fVm, EF, DF) = (AB, BC, CA). (Which property of proportionality supports this deduction?) Therefore AABC — ADEF implies that AD£F- AABC, and similarity for triangles is a symmetric relation. 404 Similarity Chapter 10 Transitive* Let it be given that AABC ~* ADEF and that ADEF — &GHL The proof that A ABC ~ AGH1 is assigned as an exercise. From tills it follows that similarity for triangles is a transitive relation, THEOREM 10,5 The perimeters of two similar polygons arc proportional to the lengths of any two corresponding sides. Proof: As in the case for Theorem 10.4, we shall prove Theorem 10.5 for triangles. It is easy to modify this proof to get a proof for quadri- laterals, pentagons, and so on, Ijet A ABC — AA'B'C be given and let p be the perimeter of A ABC and p* the perimeter of AA'B'C, By the definition of similarity we have (AB, BC, CA) = (A'B\ B'C, CA'). By Theorem 10,2, (AB, BC,AC, AB + BC+ CA) = {A'B\ B'C, A'C, A'B' + B'C + CA'). Bui p = AB + BC + CA and p* = A'B' + B'C + CA', Therefore (AB, BC, CA. p) = (A'B', B'C, CA\ p% It follows that (AB, p) = (AW, jf) and, from the Alternation and Inversion Properties that (p,p') = (AB.A'ff). In a similar way, it can be shown that (p, p') = (BC, B'C) and (p, p') f (CA, CA'). This completes the proof of Theorem 10.5 for triangles. EXERCISES 10.4 1. Complete Ihe proof of Theorem 10.4 by proving thai A ABC- AGHl 2. Given A ABC - ADEF, mLA = 30, m£B = 60, AB m 20, BC = 10, CA = 10 V3, DE = 100, find mlD, mlE, mZF, EF, and DF, 10,4 Similarities between Polygons 405 In Exercises 3-10, ihcre are figures showing two similar triangles with some side lengths indicated and a similarity statement given. In each case, find the proportionality constant of the given similarity statement and Uks lengths of any sides whose lengths are not given. 3. A ABC- 6D£P 4, A a P 7 S S q A ADC - AACB &PQR- ATQS 7, T ATSZ - AXV7 406 Similarity 8. D Chapter 10 9. J? t l A ARKD- A HAT &ABC - AA'R-'C 11. Given: A ABC <~ AA'B'C AA'B'C — AA"£"C" AB = JA'S* A'W - 5A"B" Pmce: AABC =g AA"J?"C" 12. Let AAXT — AAtt/D wilh A\ AT AT HU AfD DD = 1000. If the shortest side of AMTflD is of length 1000, find the length of the shortest side of A A AT. 10,4 Similarities between Polygons 407 13. II in Exercise IS, p and p' denote the pcriinclcrs of the smaller triangle and tile larger triangle, respectively, find *— ■ P 14. ABCDE and A'B'UD'E' are pentagons such that (AB t BC, CD, DE, EA) = (A'B\ WC t CD f t &E% E'A% £A=*£A\ LBzt£W t £Cm£O a LD^LD\ £Bat£&, and AB = ISA'S 1 . Prove that the perimeter of ABCDE h 13 times the perimeter of A'B'CD'E'. 15. If A ABC ~ AA'B'C and AB = 10, HC = 8, A'fl' = 25, A'C = 35, find B'C and AC. 16. If AFQR ~ ASTV and PQ = 24, ST = 16, PR = 18, TV = 10. find Qfl and SV. 17. If A ABC ~ ARST and 7 ■ AB = 4 « RS, what is the ratio of the per- imeter of A ABC to the perimeter of ARST? 18. If A ABC- ADEF, AB = 5, BC = 7, AC = 8, and the perimeter of ADEF is 60, find DE, £F t and DF. 19. In the figure, A-K-T-S, RK ± S3, RK = 3. AT = 8, 15 = 4. (a) Find | AHAT|, the area of A RAT. (b) Find |AflTS|, (c) Solve for at: (| A RATI, |ARIS|) = (,v, 4) 20. In the figure, A-K-T-S> M. 1 A3, JW: = 3, AT = 12, TS = 2. (a) Find [ARAT|. (b) Find ,AHTS|. (©) Solve for *: (*, ARTS|) = (12, 2). 408 Similarity Chapter 10 21. In the figure. Af-S-O-T, M J. S7T, MO = 12, OT = 5. (a) Solve for xs flAiVAfq, A.VOTJ) = (12, *). (b) Solve for tjx (\ANMO\. y) = (|A.VOT| P 5). In the figure, R-F-S-T, R-D-E-W, SE || TW t FE ± KT, D5 ± EW, RS = 20, ST = 10, RE = 30, EW = 15, FE = 29. (a) Find \ARSE\, (b) Find | AST£|. (c) Find SD, (d) Find | A SEW]. In the Bgure, A-D-B, A-E-C, Z5E || BC, £D = 12, DA \&BDE\ = 120. (a) Find;AADE|. (b) FirtdjADECj. (c) Find the ratio of AE to EC. (d) Find the ratio of AD to DB. (e) Compare the ratios in (c) and (d). 10,5 Some Length Proportionalities 409 10.5 SOME LENGTH PROPORTIONALITIES In this section there arc several theorems regarding the lengths of segments formed by lines parallel to one side of a triangle intersecting the other two sides, and there are other theorems extending these ideas to lengths of segments formed when three or more parallel lines are cut by two transversals. These theorems are useful in proving the similarity theorems of Section I0;6. THEOREM 10.6 (Triangle Proportionality Theorem) If a line parallel to one side of a triangle intersects a second side in an in- terior point dien it intersects the third side in an interior point, and the lengths of the segments formed on the second side are propor- tional to the lengths of the segments formed on the third side. Proof: Let there be given a triangle A ABC and a line / such that I intersects KB in an interior point D and such that t is parallel to EC as suggested in Figure 10-5. We know from Theorem 2.6 that Winter- sects A(7 in an interior point Call it £. We shall prove that (J5D, AD, BA) = (CE, A£, CA). A Figure 10-5 Ut hi be the distance from E to AB, /*2 be the distance from D to AC t ft 3 be die distance between t>E and BC. Following are the main steps in the deductive reasoning which com- pletes the proof of the theorem. You are asked in the Exercises to sup- ply the reasons for steps 1 through 7 (definitions, theorems, preceding steps; etc.). 1. BD ^hj-BD _ ABED\ AD ~ ^i t - AD " \AAED\ _ CE Mio-CE _ \ACED\ * AE " %hg'AM ' AAED 3. | ABED\ = ^ 3 • DE = \ACED\ 410 Simitar ity Chapter 10 4 BD = CE ' AD AE 5. BD-AE = AD'CE 6. (BD, AD) = (CE t AE ) 7. {BD, AD, BA) = (CE, AE, CA) Therefore, the lengths of KB and the segments formed by I cutting KB are proportional to the lengths of 7l€ and the segments formed by I cutting JlC. From step 7 we get step 8, 8. {BD, BA) s (CE, CA) and (AD, BA) = (AE, CA) From step 8 we got step 9 using the Alternation Property. 9. (BD, CE ) = (BA, CA) and (AD, AE) = [AB, AC) Therefore, the lengths of AB and the segments funned by I cutting other two sides in interior points, then it exits oif segments whose lengths are proportional to the lengths of those sides. THEOREM 10.7 (Convene of the Triangle Proportionality Theorem) Let A ABC with points O and E such that A-D-B and A-E-C be given. If (AD t AB)~(AE t AC), then DE I BC. Proof: Let AABC with points D and £ such that A-D-B, A-E-C, (AD t AB) =; (AE t AC), as suggested in Figure 1Q-6, be given. B Figure MMSI Let I be the unique line through D and parallel to BC. Let W be the point in which ] intersects AT. In the figure, E and A" appear to be dif- ferent points. We shall show that they are actually the same point. Since I is parallel to BC, It follows from Theorem 10-8 that (AD,AB) = (AE\AC). 10,5 Some Length Proportionalities 411 But it is given that Therefore Then (ADtAB) == (AE.AC). (AE\AC) = (A£, AC). Why? AE'-AC = AE-AC (Why?) and AE' = A£. Since E and £' arc points of AC that are on the same skle of A and at the same distance from A* it follows that £ = E'. Since / = ££' = M t it follows that BE || 5C. The word "cut" is used frequently as a synonym for "intersect/' particularly in situations involving several lines and a transversal,, as in our next theorem. THEOREM 10.8 If two distinct transversals cut three or more distinct lines that are coplanar and parallel, then the lengths of the segments formed on one transversal are proportional to the lengths of the segments formed on the other transversal. Proof: We shall prove the theorem for three p. ir.il lei lines. It is easy to modify the proof for more than three parallel lines. V Figure 10-7 Let * and / be two distinct transversals of three distinct lines (, m, n that are coplanar and parallel as suggested in Figure 10-7. Label the points of intersection as in the figure. Draw XF. Let G be the point of intersection of m with AF. It follows from the Triangle Proportion- :tlt(\ Tli or .rem. applied to ;\ACh, rhat I: {AB > AC) = (AG t AF) and from the same theorem, applied to &AFD, that (2) (GA, FA) = (ED, FD). 412 Similarity Chapter 10 From the Transitive Property of Proportionality it follows that :V3) (AB t AC) = (DE,DF). Similarly, we may show that (4) (BC t AC) = (EF,DF). Then from (3) and (4) we gel (5) (AB, BC, AC ) = (DB, EF t DF ). COROLLARY 1Q.8J If a line bisects one side of a triangle and is parallel to a second side, then it bisects the third side. Proof: Assigned as an exercise* EXERCISES 103 In Exercises 1-5, there is a triangle, &ABC t with points D and £ such that A-D-B and A-E-C, 75E | BC, and with lengths denoted as follows: p = AD, q = DB, rmAE,tm EC In each case, given three of the four numbers p. q, r, s, find the missing one. Draw and label a figure for each of these exercises, 1. p = 5, q a 6, r = 75, * = [7] 2. p = 8, q = G> r = 0, « = 9 3. p = 18, <f = (TJ, r = 10, * = 13} 4. p = [0, </ = »$, t = 7, * = 20 5. p = V£ </ = "\A *" = 4 * = In Exercises 6-10, there is a triangle, A ABC, with points and E such tliat A-IKBaudA-E-C, and with lengths denoted as follows: x = AH,jj = AC T p — AD, q = DB, r = AE, tt = EC. In each case, given some of these lengths, determine whether or not the lines DE and BC are parallel. Draw a figure for each exercise. 6. x = 10, if = is. p = 5, r=S 7. * = 10, !/ = 15, p = 5, r=7.5 8. p = 25, q = i5, r = 60, * = 3fl 9. p a 0.9, s = 0,88, r = 0,81. q = 0.81 10. p = ^ * = 2, r=2, t/ = 4 11. Write a "reason" for each of steps 1 through 7 in the proof of Theorem 10.8, 10.5 Some Length ProportDomlitit* 413 12. Coplan.tr and parallel lines a, b, c r d arc cut by transversals s and f sis •ii i'-."/-' ; Mn the figure, Given Lengths C& -> cements SS labeled in Lite lim- ine, find x and tj. _^ ■fro -M 5.4 -¥d 13. If tlirer distinct c:oplanar and parallel lines arc cut by two distinct par- allel transversals, then the lengths of the segments formed on one trans- versal are proportional to the lengths of the segments formed on the other transversal. What is the constant of proportionality in this case? 14. In the proof of Theorem 10.8, Figure 10-7 suggests that s and / do not intersect in the portion of the plane between lines / and n. Draw a figure for this theorem which shows ,v and * intersecting at a point between lines / and n. b the proof for this theorem, as given, applicable for the case suggested by your figure? 15. Consider again Figure 10-7. Suppose that the figure is modified to show s and f intersecting at a point P on the opposite side of hne / from it Using Theorem 10.6 and Figure 10-7 suitably modified and labeled* obtain some proportionalities involving lengths of segments with P as one endpoint, Use these proportionalities to prove that (AB, AQ = (DE, DP). 16. Complete the following proof of Corollary' 10.8.1. Let A ABC be given with D the midpoint of AB and let I he the line through D, parallel to BC, and Intersecting A Cat E as shown in die figure. Let l\ be the line through A and parallel to J. Then l x | BC. Why? Then {AD t DB) = (AE, EC). Why? It follows that AD_ _ AE _ , .„. „ DB ~ EC~ ' Complete the proof by showing that / bisects AG. A 4 — d/ \e Hi 1 414 Similarity Chapter 10 17. (An in of con length tractoi Copy formal geometry exercise.) With the aid of a rale i passes, draw two triangles, A ABC and AA'B'C of the sides (in centimeters) are as listed in the table. ; measure the angles A, B, C, A', W, C to the n© the table and complete it by recording the angle in r and a pair so that the Using a pro- test degree. fiisurev AABC AA'B'C AB = 9.0 A'W = 13.5 BC = 7.0 ffC = 10.5 AC = 6,0 A'C = 9.0 m£A = [£J mlB = [?J roZC = [T| m/.A' = g] mLW m E3 mlC = |1] 1& {An informal geometry exercise.) With the aid of a ruler and protrac- tor, draw two triangles, A ABC and AA'B'C, with side lengths (in centimeters) and angle measures as indicated in the table. Measure the remaining parts of the two triangles and record the results. Measure lengths to the nearest 0,1 cm. aud angles to the nearest degree. AABC AA'B'C AB = 8.0 A'B' = 10.0 BC = 10.0 JTC = 12.5 AC = CO A'C = [7j mlA = {jl m£A' = {2] mZB = 46 m/.B' = i6 i«ZC = (T) inZC = [TJ lfc {An informal geometry exercise.) With the aid of a ruler and a pro- tractor, draw two triangles* A ABC and AA'B'C with side lengths (in centimeters} and angle measures as indicated in the table. Measure the remaining parts of the two triangles and record the results. Measure lengths to the nearest 0.1 cm. and angles to the nearest degree. AABC AA'B'C' AB = 10 A'B' = 6 BC = CD B'C = \T] AC=\?] AC = mZA = 32 mZA' = 32 mZB = 51 mLW = 51 mlC=\2} mZC = [?3 10. 6 Triangle Similarity Theorems 41 5 10.6 TRIANGLE SIMILARITY THEOREMS In our study of congruence for triangles we first defined congru- ence so that, by definition, all six parts of one triangle must be congru- ent to the corresponding parts of a second triangle in order for the triangles to lie congruent to each other. On the basis of our experi- ences with triangles it seemed reasonable to expect all six of the re- quired congruences involving sides and angles to be satisfied if certain sets of three of them are satisfied So we adopted the well-known Tri- angle Congruence Postulates, referred to as S.A.S., A,S,A., and S.S.S, Similarly, our experiences with triangles, especially the triangles of Exercises 17, 18, 19 of Exercises 10,5, suggest that, if certain combina- tions of some of the definitional requirements for a triangle similarity are verified, then all of the requirements for a similarity are satisfied. Since we adopted Congruence Postulates, it would seem reasonable to adopt Similarity Postulates. It turns out, however, that it is not difficult to prove what we want to know about similarity; hence in this instance postulates are not necessary. First, we prove a theorem that is useful in proving the main Similarity Theorems. THEOREM 10.9 If A ABC is any triangle and k is any positive number, then there is a triangle AA'B'C such that AA'B'C *— A ABC with constant of proportionality k. Proof: Let triangle A ABC and a positive number k be given. We consider three cases. Case 1. k <C \. Case 2. k = L Case 3. k> 1. We shall prove the theorem for Case 1 and assign the other two Cases in the Exercises. a Suppose that k < 1; then there Av is a point D on AB such that / ^s^ AD = k* AB and a point E on p/— — .--J>. £C such that AE = k*AC In / / ^\ Figure 10-8, DE is drawn so that g f -^5, k appears to be about 0,6 , Figure io-8 It follows from the converse of the Triangle Proportionality Theorem that DE |, EC, It follows from theorems regarding parallel lines that LADE & L B and LA2ED ss L C. 416 Staiilirfty Chapter 10 Of course. LA^ I A. As you might expect, it is AADE that qualifies as it suitable AA'FC, that is, ADE «— * ABC is a similarity. So far we have shown that corresponding angles are congruent and, from the way we have chosen D and E t we know that (AD, AE) = (AB y AC). We need to show that {AD, AE, DE) = (AB, AC. BC). _Let F be the point of BC in which the line through £ and parallel toJB intersects BC. Then DE = BF(Why?) and {AE, BF) y (AC t BQ. Why? Substituting, we get (AE, DE) = (AC\ BQ. From tins proportion and the preceding proportion, (AD,AE)j(AB s Aq, it follows that (AD, DE, AE) = (AB t BC, AC), which completes the proof for Case 1 in which k < 1. THEOREM 10.10 (S,S.S. Similarity Theorem) Given A ABC and ADEF, if (AB, BC, CA) = (DE, EF, FD), then A ABC— ADEF. Proof; Let A A BC and A DEF such that (AB, BC, CA) = [DE, EF, FD) he given. (Sec Figure 10-9.) Suppose the constant of proportionality is k. From Theorem 10.9 it follows that there is a triangle AD'E'F such that AD'E'F *— ADEF Witt) proportionality constant ft. Then (AB. BC, CA) = (DE. EF, FD) with proportionality con- stant k. (&E\ FF, FD-) = (DE, EF, FD) with proportionality con- stant k, (AB,BC,CA) = (D , E\ET,F'D f ) with proportionality con- stant 1. Whv? 10.6 Triangle Similarity Theorems 417 D* *F D' Figure 10-9 From this we conclude that AB = DT', BC = E'F, and CA = Ft/. It follows from the S.S.S. Congruence Postulate that AABC== AD'E'F, BecaU now that triangle congruence is u special case of triangle simi- larity and that triangle similarity is an equivalence relation. There- fore AABC- AD'E'F. But AD'IT-ADEF, it follows thai A ABC— AD£F, THEOREM 10.11 (S.A.S. Similarity Theorem) Given AABC and ADEF, if Z A =s LD and (AB, AQ = (DE, DF), then AABC- ADEF. Proof: Let AABC and ADEF be given with L A ss Z D and (AB, AQ = (D£, DF). (Use Figure 10-9 again.) Suppose die constant of proportionality is k. Let AD'E'F be a triangle such that AD'F/F ~ ADEF with propor- tionality constant k. Then AB = fc-DE, D'£' = Jfc«D£ t AB = WB % and AC? = k ■ DF, DT = Jt ■ DF t AC = D'F. It follows from the S.A.S. Congruence Postulate that AABC = AD'E'F. Then AABC ~ AD'E'F and AUWF ~ ADEF, and wc may con- clude that AABC - ADEF. COROLLARY 10.11.1 A segment which joins the midpoints of two sides of a triangle is parallel to the third side and its length is half the length of the third side. Proof: Assigned as an exercise. and AA=*AD f ; and IB^AF/- and AB = D'E'. 418 Similarity Chapter 10 THEOREM 10.12 (A.A. Similarity Theorem) Given A ABC and ADEF, if A A s Z D and ZB ss Z E, then AABC — AJ9£F. Proofi Let AABC and ADKFsuch liiat LA m AD and AB~z AE he given. (Use Figure 10-9 once more.) Let AB _ k Let AD'E'F be a Iriangle such thai AD'E'F -* ADEF with propor- tionality constant rf, Then AA^ID, ADsiAD\ ABmAM, AEatAE\ AS = k-DE, DT = k-DE, It follows from the A.S.A. Congruence Postulate that AABCs ADTF. Then A ABC - AD'E'F and A£>E'F - ADEF, and we conclude that A ABC — ADEF. Note that we have an S.S.S. Congnience Postulate and an S.S-S. Similarity Theorem f and that we have an S.A.S. Congruence Postulate and an S.A.S. Similarity Theorem, hut that we do not have an A.S.A. Similarity "Theorem to match our A.S.A. Congruence Postulate. Of course, we could, if wc wished, call our A.A. Similarity Theorem die A.S.A. Similarity Theorem. But if Z A a* AD and AB =* AE, we do not need to be concerned, about whether "AB is proportional to DE," Indeed, if AB and DE are any two positive numbers whatsoever, there is a number k such that AB-a It -JOB. Look at the tables you prepared for Exercises 17, 18, 19 of Exercises 10.5. Do the measurement data recorded in the tables illustrate the triangle Similarity Theorems? They should. Which theorem does Excr- eta 17 illustrate? Which theorem docs Exercise 18 illustrate? Which theorem does Exercise UJ illustrate? We have written the triangle Similarity Theorems using quite a few symbols. Is it possible to state them in a more relaxed form without symbols? In the following versions of the theorems we use the word ^corresponding" without "pinning it down." It should be understood in each case that a correspondence between the vertices of one triangle and the vertices of the other triangle is fixed so that there are corre- sponding parts. 10.6 Triangle Similarity Theorems 419 THEOREM 10.10 (S.S.S. Similarity Theorem— Alternate Form) If the lengths of the sides of one triangle are proportional to the lengths of the corresponding site of the other triangle, then the triangles are similar. THEOREM 10.11 (S.A.S. Similarity Theorem— Alternate Form) Tf an angle of one triangle is congruent to an angle of another tri- angle and if the lengths of die including sides are proportional to the lengths of the corrcspondint; sit Irs in the oilier triangle, then the triangles are similar. THEOREM 10J2 {A. A. Similarity Theorem— Alternate Form) If two angles of one triangle arc congruent to the corresponding angles of another triangle, then the triangles are similar. The following theorem points out that if the sides of one triangle are parallel to the sides of a second triangle, then the two triangles are similar. The property of parallel sides is a sufficient condition to ensure similarity* but, of course, it is not a necessary condition. THEOREM10.13 U triangles AABC and ADEF are such that A~B .1 m. BC I! EF, UA II FD, then AABC - ADEF. Proof: Assigned as an exercise. EXERCISES 10.6 1. Prove Theorem 10,9 for the case in which k = L 2. Prove Theorem 10.9 for the case in which k > 1. In Exercises 3 and 4. two triangles and the lengths of their sides are given by means of a labeled figure. a b AABC — APQR? Is AABC ~ AQPR? h AABC- APRQ? Is AABC— ARPQ? Is AACB ~ ARQF? Is ACBA - AQPR? 4. Is AABC— A CDE? h AABC— ADEC? Is AABC- AEDC? Is A CAB- A DEC? Is AC/iA — ADCEr Is ABAC— A DEC? 420 Similarity Chapter 10 In Kxerciscs 5 and 6, two triangles arc given in a figure wilh some segment lengths and angle measures. 5. 6. Is AABC Is AABC Is AABC Is AABC Is Aj\BC Is AABC ADEF? A EFD? ADFE? AEDF? AFED? AFDE? Is AABC ~ ADEB? Is AABC— ADKE? Is A A KG— ABED? Is AABC— &JHBEP Is AABC— AEBD? Is AABC ~ AEDB? 7. Given isosceles AABC with AB ~ AC* and with points D, /£, Fsueh dnit A-D-B, JJ-E-C, C-F-A, DE±AB, FE 1 XC, prove that ABD£— ^C3^. 8, If at a certain Lime, in a certain place, a certain tree casts a shadow 40 ft, long and a 6-ft, man casts a shadow 2 ft. and 3 in. long, find the height of the tree. Exercises 9-11 refer to the figure with A-E-C, B-D-C, ED || A~B, and seg- ment lengths as marked. 9. Name a pair of similar triangles and explain why they are similar. 10, Find r. 11, Find* 12, In the figure, A-D-C and LABC^lBDC. Name a pair of similar triangles and explain why they are similar. 10.6 Triangte Similarity Theorems 421 D C 13. In the figure* AB || CD, IDE X AC, W±7kC t A-E-F, and E-F-C. Name a pair of similar triangles and explain why they are similar. Exercises 14-20 refer to the figure with BTJ ,| OE, DB || EC, A-B-D, D-E-F, A-C-F f and A~F X F25. 14. Prove AAFD ~~ &CFE. 15. Prove ACF£ — AACB, 16. Prove AAFD — AACB, 17. If CF = 2, BD = 3, and AC = 8, find AB. 18. If AB = 12, BD = 3, and BC = 8, find DF 19* If AD = 18, AF = 9, and AC = 7, find EC. 20. If CF = 3, KF = 4, and AC = 7, find DF. 21. In the figure, ^fE and CD are altitudes of AABC, A-F-E, and C-F-D. (a) Prove AB£A - A.BDC. (b) Prove AAW- ACEF. 22. Given AAJ3C with points D and £ such that A-D-B, B-E-Q AE X BC, CD X SB, prove that (AEtEfy-iCDrDB). 23. Given two right triangles, A ABC and AAPQ, as in die figure, copy and com- plete the following proportionality involv- ing lengths of the sides of these triangles: {AB.BC,CA) = (AP T [2],[I]). 24. Given AABC - ADEF, AB = §, BC = 7, AC = 10. DE = 7, fi EF and DF 422 Similarity Chapter 10 25. Given A ABC - APQH prove dial if A ABC is a right triangle, then APQR is also a right triangle, 26. Prove Corollary 10.11.1. 27. CHALLENGE PROBLEM, CiVCIl parallelogram ABCD with B-E-C, AE and ED intersect- ing at h\ and BF = $-BD, prove that BE = £ - BC 28. Prove Theorem 10.13. Consider two cases: (a) AB and DE are parallel, BC and EF are parallel, CA and VD are parallel, and j b) AM and 15 JS are antiparailcl, BC and /Li*' are antiparaUel, CA and F£) are an ti parallel Use Theorems 7.26 and 7.28. 10.7 SIMILARITIES IN RIGHT TRIANGLES Sometimes base and altitude are interpreted as segments and some- times as numbers (lengths of segments). In our next theorem altitudes are segments If AABC ~ AA'B'C, then we have agreed that A and A' are corresponding vertices, KB and A'B' are corresponding sides, and so on. It is natural to extend this idea to include corresponding altitudes, that is, altitudes from corresponding vertices. THEOREM 10.14 If two triangles are similar, then the lengths of any two corresponding altitudes are proportional to the lengths of any two corresponding sides, Proof: Given A ABC ~ AA'B'U, let D and D' be the feet of the altitudes from A to E€ and from A' to B'C, respectively. Let a = BC, b = CA, c = AB, h = AD, a' = B'C, b' = CA\ c' = A'B', h' = A'D'. We shall prove the theorem for the case in which B-D-C, as shown in Figure 10-10, The remainder of the proof is assigned in the Exercises . >1^ Figure 10-10 10,7 Similarities In Right Triangles 423 Since B-D-C, Z B and Z C are acute angles. Why? Then Z B' and Z C are acute angles. Why? Then it is impossible for D' to be either the point £' or the point C\ Why? Also, it is impossible that D'-B'-C or that W-C-D'. If ET-W-C, then AB'&A' is a right triangle with an acute exterior angle contrary to the Exterior Angle Theorem. There- fore B'-D-C as indicated in Figure 10.10, Now /_B ^ LB {Why?) and LBDA ^ LB'D'A* (Why?). It fol- lows from the A.A. Similarity Theorem that AADB — AA'D'B'. Therefore But (c. ft) = (c', A'). (a, 6, o) = (a', b f , c'l Therefore (a, c) == (a\ c') and (fo, cj = (//, It follows from the Alternation and Inversion Properties of Propor- tions that {K /i') f (c, v'% (c, c') = (a, a% and (c, c<) f (ft, I,*). It follows from the Equivalence Properties of Proportionalities that h and h' are proportional to the lengths of any two corresponding sides. THEOREM 10,15 If two triangles are similar, then their areas are proportional to the squares of the lengths of any two corre- sponding sides. Proof; Given A ABC ~ AA'iTC, let D and D' be the feet of the altitudes from A to W and from A' to B'C, respectively. Let a = BC, h = CA, c = AB r h = AD, a' = B'C. 1/ = CA\ tf = A'B\ h r = A'D'. (See Figure 10-11.) It follows from Theorem 10,14 that A B D B- ir Figure 10-11 424 Similarity Chapter 10 Suppose that h = ka; then h' = kif and \AABC\ = {ah = %a(ka) = $k)a* t | AA'B'C' 1 = \u'h' = \<f{W) = (JfttfA Therefore the areas of A ABC and AA'jB'C are proportional to a- unci (rt') 3 . Similarly, it may be shown that the areas are proportional to h 2 and (by and'to c 2 and (c') a . THEOREM 10.16 In any right triangle the altitude to the hypote- nuse separates the right triangle into two triangles each similar to the original triangle, and hence also to each other. Proof: Let A ABC be a right triangle with the right angle at C and with D the foot of the altitude to the hypotenuse. Then D ,/- A and D /- B. (Which theorem is the basis for this assertion?) Also* it is iin possible to have D-A-B or A-B-D. (If either of these betweenness re- lations ig true, there is a triangle with D as one vertex with one interior angle a right angle and one exterior angle an acute angle. Which the- orem does this contradict?) Therefore D is an interior point of AB as suggested in Figure 10-12. Figure 10- IS You are asked in the Exercises at the end of this section to complete the proof by showing that A ABC, AACD, and I\CBD arc all similar to each other. Next we have two corollaries that follow easily from Theorem KXIG, but first we need some definitions. Definition 10.4 If P is a point and I is a line, the projection of P on Ms (1) the point P if F is on / and (2) the foot of the perpendicular from P to I if P is not on /. Definition 10.5 The projection of a set S on a line / is the set of all points Q on t such that each Q is the projection on J of some P in S. 10.7 Similarities in Right Triangles 425 Compare Definitions 10.4 and 10.5 witb Definitions 8,9 and 8.10 in which we denned a projection on a plane. Note in Figure 10-12 that AD is the projection of AC on AB. In- deed, A is the projection of A T D is the projection of C, and every point Q such that A-Q-D is the projection of a point P such that A-P-C, Conversely, every point P such that A-P-C has as its projection on AB a point Q such that A-Q-D. Since the projection of AT on AB is a part of the hypotenuse in the situation of Theorem 10.16, we may say that AD is the projection of AC on the hypotenuse. Similarly, 7M is the projection of CB on the hypotenuse. COR OLLAR Y 10J0.1 The square of the length of the altitude to the hypotenuse of a rigfat triangle is equal to the product of the lengths of the projections of the legs on the hypotenuse. Proof; In Figure 10-12, AD is the projection of AC on AB, and UB is the projection of WC on AB. In the notation of the figure, we must prove that (CUP^AD-DB. Using Theorem 10.16 and some properties of proportionalities, wc have AACD - ACBD (AC, CD, AD) = (CB f BD, CD) {CD, AD) = (BD, CD) (CD)* = AD*DB COR OLL\R Y I 0. 1 &£ The square of the length of a I eg of a right triangle is equal to the product of the lengths of the hypotenuse and the projection of that leg on the hypotenuse. Proof- Assigned as an exercise. Definition 10.6 If a and b are positive numbers such that (a, x) = (*, b) or that fe *) f ft *). then % is called a geometric mean of a and h. 426 Similarity Note that if or if Chapter 10 (*. x) f fe b) (x, a) = (h, 4 then x 2 = ab (Why?) and ar = y/aB or x = — yS5. We often call y/aB the geometric mean of a and b. In view of Definition 10,6, Corollary 10.16 .1 and Corollary 10.16.2 can be restated as follows, COROLLARY 10.16.1 {Alternate Form) The length of the alti- tude to the hypotenuse of a right triangle is the geometric mean of the lengths of the projections of the legs on the hypotenuse. COROLLARY 10A6.2 (Alternate Form) The length of a leg of a right triangle is the geometric mean of the lengths of the hypote- nuse and the projection of that leg on the hypotenuse. In Chapter 9 we proved the Pythagorean Theorem using proper- ties of areas and suggested two other area proofs in the Exercises. One of the shortest proofs of the Pythagorean Theorem is an algebraic proof that follows easily from Corollary 10.16.2. We state the Pythagorean Theorem again and outline a proof that employs Corollary 10, 16.2, We also proved the Converse of the Pythagorean Theorem in Chapter 9. We state the converse again; we shall not prove it again. THEOREM 10,17 (TJte Pythagorean Theorem) In any right tri- angle the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs. Proof: Let A ABC With a right angle at C be given. (See Figure 10-13.) Let D he the foot of the altitude to the hypotenuse AT?. Ixt AR = c,BC = a, CA = b. AD a x t twdDB == c - *. Then it follows from Corollary 10.162, with a the length of a leg and c — x the length of its projection on the hypotenuse that a 2 = (c — x)c t and with b the length of a leg and x the length of its projection on the hypotenuse that b 2 =. xc .The proof may be completed by .showing thala? ■+ fr 2 = c 2 . Figure 10-13 A * D 10.7 Similarity* in Right Triangles 427 THEOREM 10.18 {Converse of the Pythagorean Theorem) If a 2 + fc 2 = c 2 , where a, h, c are the lengths of the sides of a triangle, then the triangle is a right triangle with c the length of I he hypotenuse. EXKKCJSES 10.7 I, Given right triangle A ABC with hypotenuse AB, let D he the foot of the altitude to EC. E the foot of the altitude to AC, and P the foot of the altitude to AB, How many distinct points are there in the set (A,B, C,«, E.F)? In Exercises 2-7, there is a figure show mga right triangle with hypotenuse AT3 and with D the foot of the altitude to aB, In each case, given the lengths of some of the six segments, find the lengths of the other segments. Express your answers in exact form using radicals if necessary. 3. B C 4 4, b 428 Similarity Chapter 10 H, Find the perimeter of an equilateral triangle if the length of each of its altitudes is 10. 9. Find the length of the diagonal of a rectangular floor to the nearest foot if the floor is 21 ft. wide and 28 ft. long. 10. A ladder 12 ft long reaches to a window sill on the side of a house. If the window sill is 9 ft. at»ove the (level} ground, how far is the foot of the ladder from the side of the house? 11. Find the length of the hypotenuse of a right triangle if its legs each have length 1. Express the answer exactly using a radical. 12. Find the length of the hypotenuse of a right triangle if its legs each have length 100. 13. Find the lengths of the legs of an isosceles right triangle whose hypote- nuse has length 1. 14. Find the lengths of the legs of an isosceles right triangle whose hypote- nuse has length 2. 15. Find the lcugtlis of the legs of an isosceles right triangle if the length of its hypotenuse is \/2. 16. If one leg and the hypotenuse of a right triangle have lengths 1 and 2, respectively, find the length of the other leg. 17. Find the I ength of a leg of a right triangle if the other leg and the hypote- nuse have lengths 1 and \/3~, respectively. 18. Find the length of the leg of a rigfrl triangle if the other leg and the hy- potenuse have lengths 100 and 100\/3~ t respectively. 19. Given A ABC with m£ C - 90 and with D the midpoint of AS aiid£ the midpoint of EC, prove that ACED m ABED. 20. For AA BC t m Z C = 90 and D is the midpoint oiA~B. If AC = y7 and BC = 3, find CD. 21. Complete the proof of Theorem 10.16 by showing that ABC «— * ACD and ABC * — ► CBD are similarities. (See Figure 10-12.) It will then follow from the equiva- lence properties of similarity for triangles that ACD * — * CBD is also a similarity. 22. Prove Corollary 10.16.2 fOT the leg AC in Figure 10-12. 23. Prove Corollary 1Q.1&2 for the leg EC in Figure 10-12. 24. See the proof of Theorem 10.17. Show that a 3 + fe* = c 2 . (See Figure 10-13.} 25. If A ABC ~ A DEF and 5 • A = 3 * DE, what is the ratio of the length of an altitude of the smaller triangle to the length of the corresponding altitude of the larger triangle? Which theorem justifies your answer? 26. In Kxcrcise 25, what is the ratio of the area of the smaller triangle to the area of the larger triangle? Which theorem justifies your answer? 10.8 Some Right Triangle Theorems 429 27« If 16- \APQR\ = 25 ■ A ABC and If APQR — AAHG, what is the ratio of PR to AC? Which theorem justifies your answer? 28. Prove Theorem 10.14 for the case in which D = B or D = C 29, Prove Theorem 10.14 for the case in which D-B-C (the proof for the case in which B-C-D Is similar to the proof for the case in which D-B-C). 10.8 SOME RIGHT TRIANGLE THEOREMS Following arc some theorems regarding right triangles. Although the) f are not profound, they arc useful theorems that every mathemat- ics student who has studied formal geometry ought to know. These theorems should not surprise you. If you worked the exercises in Ex- ercises 10.7, you will recognize them as "old stuff." THEOREM 10,19 The median to the hypotenuse of a right tri- angle is one-half as long as the hypotenuse. Proof; Let AABC be a right triangle with D the midpoint of the hypotenuse* {See Figure 10-14.) We want to prove that CD = \*AB, or equivalent! v, that CD = DB. I^et point E be the midpoint of CB. Then C B Figure 10*14 (BD, BE) = (BA, BC) (Why?) and LB^LB. It follows from the S.A.S, Similarity Theorem that A ABC— A DUE. Then mLBED = mLBCA = 90 = mLCED. It follows from the S.A.S. Congruence Postulate that ACDE=* ABDE and therefore CD = DB. There are some special right triangles that are referred to in special ways. First we mention the 3, 4, 5 triangles. A triangle whose sides have lengths 3, 4 t 5 is a right triangle. We know this since 3* + 4* m 5 2 < (Arc we using the Pythagoreun Theorem when we muke tills em illu- sion, or are we using its converse?) Given a distance function, there are infinitely many 3, 4, 5 right triangles. Indeed, if A is any point in space (infinitely many choices here) and if B is any point such that AB = 5 (infinitely many choices here), there arc infinitely many possible points C so that A ABC is a right triangle with AB = 5, BC = 3 t and CA = 4, and infinitely many 430 Similarity Chapter 10 possible points C such that AB = 5, BC = 4, and CA — 3, But there are many, many more, not included among these, that are also fre- quently referred to as 3, 4, 5 right triangles as our next theorem suggests. THEOREM 10.2(1 (The 3, 4> S Theorem) If a: is any positive num- ber, then every triangle with side lengths 3x, 4x, 5.r is a right triangle. Proof: Let &ABC be a triangle with BC = 3, CA = 4, AB = 5. Let x be any positive number. Let &A'B'C f be any triangle with B'C = 3x, CA f =: 4s, A'B' = 5x. Then AA'B'C - AABC. (Which triangle Similarity Theorem do we use in making this deduction?) Since A ABC is a right triangle, it foDows that A A 'B'C is a right triangle, and this completes the proof. A triangle is called a 3, 4, 5 triangle if its sides arc of lengths 3, 4, 5 or if the lengths of its sides are proportional to 3, 4, 5, AH 3,4, 5 triangles arc right triangles. THEOREM 10.21 (The 5, 12, i.1 Theorem) If x is a positive num. ber and if the lengths of the sides of a triangle are ox, 12x, and 13x, then the triangle is a right triangle. Proof: Assigned as an exercise. A triangle is called a 5, 12, 13 triangle if the lengths of its sides are proportional to 5, 12. 13. THEOREM 10.22 (The /, J, \/2 Theorem) If the lengdis of the sides of a triangle are proportional to 1. 1, y/S. then the triangle is an isosceles right triangle. Proof: Let the lengths of the sides of a triangle be o, fc, c and suppose {«, h c) = (1, 1, Vl). Then there is a positive number k such that a = k • 1, h = k • 1, and Gssk*y/£. Then a = b, t& + 6 s = fe 2 + Jt= = 2fc 2 , c 2 = (k^f = 2&, a 2 + fc 2 = cK Therefore the triangle is an isosceles right triangle. 10,8 Some Right Triangle Theorems 431 A triangle is called a I, 1, V5 triangle, or a 45, 45, 90 triangle, if the lengths of its sides are proportional to 1 , 1 , -\/S, THEOREM 10.23 (The 1, y/3, 2 Iheorem) If the lengths of the sides of a triangle are proportiona] to 1 1 \/3, 2, then it is a right tri- angle with its shortest side half as long as its hypotenuse. Proof: Assigned as an exercise. A triangle is culled a 1, \/S, 2 triangle if the lengths of its sides are proportional to 1, y/3 t 2, A triangle is called a 30, 60, 90 triangle if the measures of it* acute angles are 30 and 60. THEOREM 10,24 A triangle is a 30, 60, 90 triangle if and only if it is a 1, \/3, 2 triangle with the shortest side opposite the 30 de- gree angle. Proof: Let A A BC be a 1 , \/3 f 2 triangle an d k a positive mm Vie r such that AC = fc, BC = s/Sk, and AB = 2k. (See Figure 10-15.) Let D he the point on opp CA such that CD — k. Then 1. AABC=z ADBC 1. Why? 2. AB = DB = DA 2, Why? 3. mlABD + mlBDA + mlDAB = 180 3. Why? 4. mlABD = mlBDA = m/.DAB = 60 4. Why? 5. ml ABC = mlDBC 5. Why? 6. ml ABC + mlDBC = 60 6. Why? 7. ml. ABC =30 7. Why? 8. ml8AC = 60 8. Why? 9. mlACB = 90 9. Wiry? Since L ABC is opposite the shortest side of A ABC, this completes the "if* part of the proof. 432 Similarity Chapter 10 Suppose next that A ABC is a 30, 60, 90 triangle. (See Figure 10-16.) Let D be the unique point on opp CB such that CB = CD, Figure 10-lfl Draw DA. Then 10. AABC s AADC 11. m/LCAB = mZ. CAD = 30 12. mLBAD as mZADB = mlDBA = GO 13. BA = AD = DB 14. BC = CD 15. BC + CD = BD 16. 2BC a AB Let BC = fc. Then 17. AB = 2k 18. (AQ* + (BC) 2 = (AB)* 19. (Aqz + A' 2 = 4*2 20. (AC)* = 3Jfc* 21. AC= VSJfc 22. (BC,CA.AB) = (1, x/3',2) 10. Why? 11. Why? 12. Why? 13. Why? 14. Why? 15. Why? 16. Why? 17. Why? 18. Why? 19. Why? 20. Why? 21. Why? 22. Whv? This shows that A ABC is a 1, \/5, 2 triangle and hence the "only if* part of the proof is completed. Note that in some of these names for special triangles the numbers are side lengths (or numbers proportional to them), whereas in others they are angle measures. There should be no confusion in regard to the 30, 60, 90 name and the 45 f 45, 90 name. Because 30, 60, 90 are not the lengths of the sides of any triangle, and 45, 45, 90 are not the lengths of the sides of any triangle. Which postulate justifies this statement? L0.8 Some Right Triangle Theorems 433 EXERCISES 10.8 L If AABC is a right triangle with ml C = 90, AC = 60, BC - 80, and with D the midpoint of AB. find CD. 2, (See Figure 10-14) Z ACS and Z DEB are congruent angles in the sit- uation represented by this figure. Find several other pairs of congruent angles. (Six more pairs would be rather good.) 3. In the proof of Theorem 10,19 we asserted that ACD£a ABDE. Write a two-column proof for this deduction. 4* In a book on the history of mathematics find something about the rope stretchers in ancient Egypt Explain the connection between rope stretchers and right triangles, 5. A baseball diamond is u square whose sides are 90 ft. long. What is the distance (to the nearest foot) between first and third base*? 6. The figure represents a cube whose six faces are 1 in. by 1 in. squares. Using the Pythagorean Theorem twice, once on ABCD and once on AABD, find AD, Express the answer exactly using a radical if neces- sary. (Why is LABD a right angle?) 7. A room in the shape of a rectangular box is 15 ft wide, 1 8 ft. long, and 8 ft. high. Find the distance to the nearest foot between one corner of the floor and a diagonally opposite comer of the ceiling. In Exercises 8-16, the lengths of the hypotenuse and one leg t>i' a right tri- angle are given. In each exercise, the triangle is a 1. l t \/2 triangle, or a 1, \/3» 2 triangle, or a 3, 4, 5 triangle, or a 5, 12, 13 triangle. Determine which one. 8. 100,50 9. 100,60 10. 100. 50 VI 11, 100,80 12, ]00, 50\/3 13, NX), 92^ 14. \/^,iv^ 15. 145, 116 16. 63, 25 434 Similarity Chapter 10 17. I .et^ A ABC be a right triangle with the right angle at C The midpoint of AB is the center of a circle which lies in the plane of the triangle and which contains the points A and B. Does the point C lie inside of the circle, on the circle, or outside of the circle? Why? The figure suggests a point A on a high bluff above a level plane. If the angle of elevation of the point A from the point C is n 90 degree angle and if it is 200 ft. from C to B t wliat is the height BA to the nearest 10 ft,? Assume that / CBA is a right angle. -::iC^S, :*>"" ~ 19, The figure represents an observer A in an airplane 5000 ft. directly above a point 8 on the ground. If J3 T C T D arc three colllnear points on the ground and if m L ABC = 90, m L BAC = 45, m L CAD s 15, find to the nearest 100 ft. the distance from C to D. 20. Find three positive integers, a t b t c such that \/a, \/' , » V^&re the lengths of the sides of a right triangle. How many such triples of positive integers arc there? 21. If rand y are any positive integers, distinct or not, show that y/x, \/y, \fx + tj are the lengths of the sides of a right triangle. 22. In the figure is shown a right triangle A ABC with CD the altitude to the hypotenuse. If AD = 1, DC = r, BD = y, show that * = y/y. 10,8 Some Right Triangle Theorems 435 23. In the figure below, ABCD is a parallelogram with AB = 76, AD m 50, m£A = 30, and h the length of the altitude from D to AS, Find \ABCD\. 24. A parallelogram has adjacent sides of lengths 22 and 14. If the measure of one of its angles is 30. find the area of the parallelogram. 25. Find the area of a rhombus of side length 12 if one of its angles has a measure of 60, 2& The measure of each base angle of an isosceles triangle is 30 and each of the two congruent sides has length 24. (a) How long is the base? (b) What is the area of die triangle? 27. In the figure. ABCD is a trapezoid with AB \TJD, AD = BC = 20, CD = 28, and ml A = mlH = GO. Find \ABCD\. Use the figure to complete the proof that in a 30, fiO right triangle the side Opposite the 30 degree angle is one-half as long as the hypotenuse. Proof: In the figure, ml A = 30, m L B = 60, and D is the midpoint of AB. Show that A BCD i,s equilateral and that BC = CD a 1 • AB. 2d. In a 30. 60 right triangle, the length of the hypotenuse is 8 V^- (a) Find the length of the shorter leg. (b) Find the length of the longer leg. (c) Find the area of the triangle. 30, Prove Theorem 10.21. 436 Similarity Chapter 10 31. Prove Theorem 10.23. 32. If u and c are positive integers such that u > v, and if A ? B t C arc points such dial AC = 2av, BC = u* - p 2 , AB = u* + &, prove that &ARC is a right triangle. (This exercise also appears in Chapter 9, but it is good for a repeat appearance here.) 33. See Exercise 32, If u uiid t: are relatively prime positive integers (this means tluit no positive integer except 1 divides both of them), if it and o are not both odd, and if u > c, then it can be shown that the three integers, u 2 — c 2 , 2uc„ i* 2 + v' £ , are relatively prime. If the lengths of the sides of a light triangle are relatively prime positive integers, the tri- angle is called a primitive Pythagorean triangle and the triple of its side lengths is called a primitive Pythagorean triple. Two examples of primitive Pythagorean triples are '(3,4,5) and (5,12,13). Find five more primitive Pythagorean triples, 34. challenge problem, TTie figure shows a right triangle A ABC with CD the altitude to the hypotenuse. If AC = p, BC = q t CD = f, prove that Jj- + i = \. CHAPTER SUMMARY The central theme of this chapter is SIMILARITY. The concept of similarity is based on our experiences with objects which have the same shape. The relationship of lengths in one figure to the corresponding lengths in a similar figure suggests the idea of a PROPORTIONALITY. In this chap- ter we studied the properties of proportionalities and we used them in de- veloping the geometry of similar pjlygons. The key theorems of this chapter include THE TRIANGLE PROPOR- TIONALITY THEOREM. THE CONVERSE OF THE TRIANGLE PRO PORTIONALITY THEOREM, THE &&& SIMILARITY THEOREM, THE S.A.S, SIMILARITY THEOREM, THE A.A. SIMILARITY THE- OREM, THE PYTHAGOREAN THEOREM, and THE CONVERSE OF THE PYTHAGOREAN THEOREM. The chapter concludes with a study of special right triangles. A knowl- edge of these triangles will prove useful as you continue your study of mathematics* Review Exercises 437 REVIEW EXERCISES In Exercises 1-1 0, complete the statement so that it will be a proportionality. 1.(5.12) = (35,0) 2. (1,2,3,4, 5) = (0,0 EJ. Q 15) 3. (25, 60, 65) = (Q], 12,0) 4. a io f 2i) = (a. mm) 5. (4. 10, 21) = ([3,6, HI) 6. (4, 10,21)= (0, 0,6) 7. (5000, ,3000, 1500) = (200, 0, 0) MiOO,400,5W) = (II],m,10) 9. (27, 27,81) = (a [7], 3) 10. (357, 1309, 833) = ([?}, 11, [7J) In Exercises 1 1-20, determine If the given statement is true or if it is false. 11. If x = y, then (5, x) - (5, y). 12. lfxyt=y, then (5, x) = (5, ij). 14If f = f'**>(3.3)f (*>!/>> 15. If^^^.aieniAx)^^/ 16. If x* = a&. then (a> x) = (*, £>), 17. If x* = ab, then (*, a) = (b, x). 18. If («. fc) = (c, <f), then (a. h) = [d, c), 19. If (a, fc) = (c, a*), then (a, <:} = (fe, rf). 20. If acf = hc t then (n, &) = (c, 5). 21. If AABC is any triangle, then AABC — AABC. Which property of an equivalence relation docs this illustrate: the Reflexive Property, the Symmetric Property, or the Transitive Property? 22. If A ABC- A DICE then &DEF ~~ A ABC. Which property of an equivalence relation does this illustrate? 23. If AABC- ADKF and ADEF — AG//I, then A ABC— ACHL Which property of an equivalence relation does this illustrate? 24. State the Triangle Proportionality Theorem and show how it may be used to prove the theorem regarding the lengths of segments formed by two transversals cutting three or more coplanar and parallel lines. 25. State the three triangle Similarity Theorems. 438 Similarity Chapter 10 2& According to Theorem 10,9, if A ABC is any triangle and k is any posi- tive number, then there is a triangle &DEF such that ADEF= A ABC with proportionality constant t. F.xplain how tills theorem was used in proving the triangle Similarity Theorems, 27. Explain why there is an A. A, Similarity Theorem but no A.5.A. Simi- larity Theorem. 2S. Prove that the altitude to the hypotenuse of a right triangle determines two triangles that are similar to each other. 29. Using similar triangles, prove the Pythagorean Theorem. 30. If A, B> C, D are points such that A-D-B, AD = BD = 25, AC = 30, BC = 40, find CD, 31. If the lengths of two sides of a right triangle are 10 and 15, find the length of the third side. (Two possibilities.) 32. Find the measures of the angles of a triangle if the lengths of its sides are \/2~. V? 2 - 33. Find the measures of the angles of a triangle if the lengths of its sides are \/S, 2\/3~.3- 34. We know that SO, 60, 90 cannot be the lengths of the sides of a triangle. Which postulate justifies this statement? ■ In Exercises 35-43, refer to Figure 10-17 in which A ABC is a right triangle with the right angle at C, CD is the altitude from C to the hypotenuse, AD = x, DB = «, AB = c,BC - a, AC = h, and CD = h. Figure 10-17 35. A ACS — A ADC and A ACS ~- A[TJ. By the symmetric and prop- erties of similarity for triangles, A[T] — A(7J. 36. h 2 equals the product of x and |T] (o, b t or tj). 37. h 2 equals the product of x and |T| {a, c, or tj), 38* a 2 equals the product of y and \T\ (k c > or x}> 39. If x = 16 and tj = 9, find h,a, and h. 40. If D is the midpoint of AS, then CD = [T] (in terms of c), 41. If m L A = 30 and c = 15, then a = [TJ. 42. If a s 10 and c = 20, then b = [U and m£A = [?]. 43. If ml A ^ 45 and a = 12, then b = JT] and c = [?]. Review Exercis«f 439 44. Copy and complete: In a 30, 60 right triangle, the side opposite the 30 degree angle is [?] (in terms of the hypotenuse). 45. If a boy 5 ft, tall casts a shadow 2 ft. long, how high (to the nearest 10 ft.) is a tree if its shadow is 73 ft. long? What assumptions did you make in working this exercise? 48. Find the distance from C to A& if AC = 10, BC = 10 V5, AB = 20, 47. If the hypotenuse and a leg of a right triangle have lengths 241 and 230, respectively, find the length of the other leg. 48. challenge froblem. Given rectangle ADEH and points B and C on AD such that HA = AB = BC = CD = DE = 1, prove that m/LEAD = mlEBD = mAECD. Chapter Joyce fi. Wtlsoii/Photo Researcher* Coordinates in a Plane 11.1 INTRODUCTION In Chapter 3 we introduced the idea of a coordinate system on a line. Recall that if F and Q are any two distinct points* then there is a unique coordinate system on FQ with P as origin and Q as unit point. Tims a coordinate system on u line is determined by choosing any two distinct points on the line, one of them the origin and the other the \ I j ii I point If, on line Z, F is the origin and Q is the unit point, then FQ is called the unit segment for the coordinate system on / determined by P and Q. A coordinate system on I is a one-to-one correspondence be- tween the set of all points of I and the set of all real numbers. The numbers associated with the points of /are called coordinates, and they can be used to determine distances (in the system based on FQ as the unit segment) between points on I. In diis chapter we introduce the idea of a coordinate system in a plane. In a plane, each point is associated with a pair of numbers, rather than a single number. After proving some basic theorems concerning a coordinate system in a plane, we develop some equations for a line. We then show bow coordinates can be used to provide simpler proofs of some geometric theorems. 4.42 Coordinates in a Plane Ctiapter 11 H.2 A COORDINATE SYSTEM IN A PLANE Suppose that a plane is given and, unless we specify otherwise, that all sets of points under consideration are subsets of this plane. Suppose further than a unit segment is given and that all distances are relative to this unit segment unless otherwise indicated. Let OX and OY be perpendicular lines in the plane intersecting in the point O as shown in Figure 11-1. Let / and / be points on OX and Vi t respectively, such that Ol = OJ = 1. There is a unique coordinate system on OX with origin O arid unit point I, This is called the \ -coordinate system, and the coordinate of a point R of OX in this system is called the x-eoordiiuiteor abscissa of R. t Y 6 i -6 -s -4 -3-2-1 -•1 /, It X 12 3 4 5 Figme 11-1 In Figure 11-1, the abscissa of R is 5. There is a unujue coordinate system on OY with origin and unit point J, This is (railed the y-eoordinate system, and the coordinate of a point S of OY in this sys- tem is called the y-coordinate or ordinateof S. In Figure 11-1, the ordi- nate of S is —2. Name the coordinates of Tand V in Figure 11-L Is it necessary to specify the coordinate system in each case? Why? The line OX is called the ■ mlnmrl OY is called the y-axiv Together they are called the coordinate axes. Their point of intersection, O. is called the origin and the plane is called the .vv-plimc. Although we usually represent a line with a segment and an arrowhead at each end, it is common practice to represent an axis with a segment having an arrowhead only on the end that "points in the positive direction." In many of the figures in this book axes are represented in this way. 11.2 A Coordinate System in a Plan* 443 The projection of a point P on a line f is (1) P itself if F is on /, and (2) the foot of the perpendicular from F to /if ? is not on /. (See Defini- tion 10.4.) Since the perpendicular segment from an externa! point to u Line is unique, each point in a plane has a unique projection on a given line in that plane. In Figure 11-2, PA is perpendicular to the x-axis at A and FB is perpendicular to the {/-axis at B. Therefore A is the projection of Fori the x-axis and B is the projection of P on the (/-axis. If 4 and 3 are the coordinates of A and B> respectively, then we call the ordered pah* of numbers (4, 3) the coordinates of F. . ^ s 4 * * a i ^ -±A s P5-4-3-2-: I 2 i 'A 5 - : . L.J Figure 11-S More generally, if F is any point in the xy-planc, the x-coordinatc (abscissa) of F is the x-coordinate of the projection of F on the x-axis. The ^-coordinate (ordinate) of F is the (/-coordinate of the projection of F on the {/-axis. We call the x-coordinate of F and the incoordinate of F the coordinates of P. The xy-coordinates or, simply, the coordi- nates of F are an ordered pair of real numbers in which the abscissa is the first number of the pair and the ordinate is the second. Thus if the abscissa of F is a and the ordinate of F is 6, then the x {/-coordinates of P are written as (a, b). THEOREM 11,1 The correspondence which matches each point in the .riy-plane with its xy-coordinates is a one-to-one correspond- ence between the set of all ordered pairs of real numbers and the set of all points in the xy-plane. Proof: In the xt/-plane, let an x-coordinate system on the x-axis and a {/-coordinate system on the {/-axis be given. Let F be any point in the given ary-plane, If A and /? are the projections of f on the x-axis and the 44* Coordinates in a Plane Chapter 11 (f-axis, respectively, let the abscissa of A be a and the ordinate of B f*? k Since each point on the *-axis has a unique x-coordinate and each point on the y-axis has a unique (/-coordinate, and since the projections of P on the x- and (/-axes are unique, it follows thai there is exactly one ordered pair of real numbers (a, h) that corresponds to the point P. Conversely, let (a, b) be any ordered pair of real numbers; then tiiorc is a unique point A on the x-axis with abscissa a and a unique point B on the (/-axis with ordinate h Also, there is a unique line fa through A and perpendicular to the x-axis and a unique line fe through B and perpendicular to the (/-axis. These two Hues intersect (Why?) in a unique point P. Hence every ordered pair of real numbers corre- sponds to exactly one point in the given xt/-plane and the proof is complete. Definition 11.1 The one-to-one correspondence between the set of all points in an xi/-plane and the set of all ordered pairs of real numbers in which each point P in the plane cor- responds to the ordered pair [a, b), in which a is the x-coordi- nate of P and b is the y-coordinate of F, is an v l i: : mate system. Since there are many pairs of perpendicular lines in a plane and since any pair of such lines may serve as axes, it follows that there are many ^-coordinate systems in a given plane. In a given problem sit- uation, we arc free to choose whichever coordinate system seems most appropriate. In view of the one-to-one correspondence between the set of alt ordered pairs of real numbers and the set of all points in a given xf/-plane, it is clear that symbols used to denote the ordered pairs may be used to denote the corresponding points. Thus, if P is the point whose coordinates are the ordered pair (4. — 3) t we may speak of the point (4, —3) or we may write F— (4, —3). Sometimes we simply write P{4, -3). It should be noted that the numbers in an ordered pair need not be distinct. Thxis (3, 3) is an ordered pair of real numbers, Of course, the ordered pair (3, 5) is not the same as the ordered pair (5, 3). Indeed, (0, b) = {e. d ) if and only if a = c and h = d. As shown in Figure 1 1-3, it is customary to think of the unit point 1 as lying to the "right" of the origin (that is, on ray OX in the figure) and of unit point 1 as lying "above" the origin (that is, on ray OY). This means, then, tin at the points on the x-axis with positive abscissas lie to the right of the origin and those points on the x-axfs with negative 1 1*2 A Coordinate System In a Plant 445 A 'V rS *? x i -3 5 ngmm 1 t<3 abscissas lie to the "left" of the origin (that is., on opp OX). Where do the points on the y-axis with positive ordinate* lie? Where do the points on the y-axis with negative ordiriates lie? There are situations in which it is convenient to think of the posi- tive part of the i-axis as extending to the left, or the positive part of the y-axis as extending downward, or some other variation. However, it will not be necessary to do this. A line in a plane separates the points of the plane not on the line into two half planes, the line being the edge of each halfplane. Simi- larly, the coordinate axes separate the points of an xy-plune not on the axes into four "quarter-planes," or quadrants, the union of whose edges is the axes. For convenience, these quadrants are numbered 1, 1 J, ITT, IV as indicated in Figure 1 1 -4 . +y a *<0 y>0 x>Q y>0 x \ in IV x<0 *>0 >'<0 Figure li-i Quadrant I is the set of all points {x t tj) such that % > and y > 0. Quadrant II is the set of all points {x t y) such that x < and y > 0, Quadrant III is the set of all points {x, y) such that X < and y < 0, Quadrant IV is the set of all points (r. y) such that x > and y < 0. 446 Coordinates in a Plane Chapter 11 We can describe the coordinates of those points (x, y) on OX by x > and y = 0, and the coordinates of those points on opp OX by x < and y = 0. Describe, in a similar way, the coordinates of those points (i% y) on OY: on opp OY, Since we usually think of an xi/-eoordinate system oriented as we have shown in Figures 11-1 through 11^1, it is customary to call all lines parallel to OY vertical lines. Similarly, we cal all lines parallel to OX horizontal lines. It is often convenient to use **above," "below," "right," "left" to describe the position of a point. However, we can get along without these words if challenged to do so. For example, we could describe the position of the point P = (2, —5) by saying that P is 5 units "below" tlie x-axis and 2 units to the "right" of the y-axis in an xiy-pl&ne. Or we could say that V is in the fourth quadrant, that it is on a vortical line which intersects the x-axis in a point 2 units from the origin, and diat it is on a horizontal line which intersects the y-axis in a point 5 units from the origin. EXERCISES 11.2 ■ In Exercises 1-S, name the quadrant in which the paint lies. h (-2,4) 5. {*, -2) 1 (7, -3) 6. (-7.3, -1) & (-V2, -5) 7. (-V5, V3) •I) 4. (1.2, 6) 8. & -I) In Exercises 9-18, describe the set of all points (x, y) which satisfy the given conditions. 9. x <C 0, y < 14. x is any real number, y = 4 10. x > 0, y < 15. x = —2, y is any real number 1L x < 0, y = 16. x > 3 S y = -5 12. x = 0, y > 17. x = 2, y < 1 13. x = 0, y < 18. xy = 19. If it = {-3, 7), and (a) if S is the point where the vertical line through H intersects the x-axis. what is the abscissa of S? The ordinate of S? What are the coordinates of S? (I) if T is the point whore ihc horizontal line through /{ intersects the y-axis, what is the abscissa of 7? The ordinate of T? What are the coordinates of 2*? 11.3 Graphs In a Plane 447 20. Of the following points, 8nd three that are eollinear: (3, -5), (5* 7), (—5, —5), .-5, 2), (ff, —5)* 21. Describe the set of all points in the xj/*plaiie for which the abscissa is —2; for which the ordinate is 6. Describe the intersection of these two set\ 22. Describe the set of all points in the xi/plane for which the abscissa is zero; the ordinate is zero, Describe the intersection of these two sets. Describe the union of these two sets. 11.3 GRAPHS IN A PLANE A graph is a set of points. To draw a graph or to plot a graph is to draw a picture that suggests which points belong to the graph. The picture of a graph shows the axes, but they are not usually a part of the graph. Of course, a subset of the axes is often a part of a graph. It is customary to label the x- and y-axes as shown in Figure 1 1-5. It is usually desirable to label at least one point (other tlian the origin) on the x-axis with its x-coordinate and at least one point (other Ulan the origin) on the y-axis with its (/-coordinate. * 4 3 2 : ■ * fc -4-3- 1 2 3 4 _1 Li Figure 11-5 In setting up an xy-coordinate system we start with three distinct points O, /, / such that 57 X 0?and Ot =_0/ = 1. as in Figure 11-1. Although it is understood that Ol and OJ arc congruent segments (based on lengths in the distance system that we consider to be fixed), it is sometimes helpful, particularly in applied problems, to take points T and / so that IT! and OJ "appear" to be different in length. If this is done, a picture of the xt/-plane may be described by saying that the "scale" on the x-axis is different from the "scale" on the y-axis. The word scale, as used here, is not part of gilt formal geometry. As far as our formal geometry is concerned the distance from O to I is the same as the distance from O to / regardless of appearance. 448 Coordinates in a Plane Chapter 1 1 If the scale on the x-axis is different from the scale on the y-axis, a graph may appeal 1 distorted. An answer to the question "When is a square not a square?" might be "When different scales for the & and y-axes are used in graphing its length and width/' For example, Figure il-6 shows a picture of a quadrilateral ABCD all of whose angles are right angles. Since AB = CD = 4 and AD a BC = 4 in our formal geometry, it is true that ABCD is a square in our formal geometry. In physical (Informal) geometry, if we were to measure the sides of quad- rilateral ABCD with a ruler, wo would find that Ihey are of unequal length and conclude diat ABCD is not a square . Figure 11-6 t *v 1 — 12 WIO cm 10) ■ r T 1 4 Ai2 t 6) fltf.G) ^ 11 12 3 4 5 6 1 ! 11 If a graph contains only a few points, it may be desirable to write the coordinates of each point beside the dot that represents iL Example 1 Plot the points A(-4, 3), B(0, -5), C{5, -2), D(4, 0). (Sec Figure 11-7,) *y 1 7]" *i 4,3 ) 0(4,0) *l -3 Li 3 I'll -2) -,5 9 %-fy .- Hxwe 11-7 Sometimes a small open circle is used to indicate that the endpoint of a segment or of a ray does not belong to the graph. In this connec- tion recall the symbol for a halflme introduced in Chapter 2. 11.3 Graphs in a Plane 449 Example 2 Draw die graph of {x t y) ; x < 2 and y = 3}. (Sec Figure 4 'V ! 2 x k. -; J 3 r Figure 11-S If there are infinitely many points in the graph, the picture may contain segments or curves, and sometimes shaded regions or arrows, to indicate which points belong to the graph. Example 3 Draw the graph of {fr y) : 1 < x < 3 or 2 < y < 4}. (See Figure 11-9.) >5 ■IE IE H t f Urn , fc u «.- L : «-' — - if — — " "* m 2 1 i -5 I * -% " , . i 1 1 Figure 11-9 Note that in Figure 11-9 we have shown l± and h as solid lines to indicate that they are a part of the graph. We have shown parts of mi and J7»2 as dashed lines to indicate that these parts do not belong to the graph. Of course, segment AD on mj and segment BC on m 2 are part of the graph. It is desirable to indicate all of lines vi] and m% in some manner since they are a part of the "Iwundary" of the graph. Let us agree, then, that if a line, a segment, or a ray is not part of a graph, but serves as a boundary to a graph , it will be shown in the graph as a dashed line, segment, or ray. Write the coordinates of the point of intersection of lines 1% and m-i in Figure 11-9. Is this point a point of the graph? 450 Coordinates in a Plane Example 4 Daw the Chapter 11 of {{x, |f) :*< -2}. (See Figure 11-10.) Figure 11-10 As Figure 11-10 suggests, the graph of {(*, y) : x < -2} is the half plane on the left of the vertical line /. The interiors of the rays shown in the graph are intended to indicate this halfplane. Why does line I appear as a dashed line in this graph? Since line I is a vertical line, it is parallel to the y-axis. Therefore all the points of line I are on the same side of the y-axis in the xy-plane. Another way of describing the graph in Figure 11-10, then, is as the halfplane with edge / on the op- posite side of line I from the y-axfc. Since line / represents the set of all points hi the ,ri/-plane whose abscissa is — 2, another way of describing line / is the line with an equation x = — 2, The halfplane pictured in Figure 11-10 is the set of all points (x, y) such that x < — 2; or we may describe it as all of the *r/-plane which lies to the left of the tine x= -2. EXERCISES 11.3 1. If A = (-2, 2), B = (3, 3), C = (4, -2), D = (-3, -3), draw the graph of the set of all points winch belong to the polygon A BCD. 2. Draw the graph of the set of points that belong to the interior of the polygon in Exercise 1 . 3. In your graph for Exercise 2, should the segments 7iB, T5C, CD, DA~ ap- pear as dashed lines or as solid lines? 4. Draw the graph of {(x, y) : x = 3 and — 1 < y < 2}. 5. Is the point (3. 2) part of the graph for Exercise 4? Is the point (3, — 1) part of the graph? 11,3 Graphs in * Plan* 451 hi Kxercises 13, trvsph lllfi led of all pimiK ;>', (/) in un Xgf pkne satisfying die given conditions. Then describe the graph of the set in words. Exercise 6 has been worked as a sample. 6. 2 < x < fi and y = 3. Graph: ' * it&3j 5(6,3) 2 x t -2 2 n 1 Figure 11-11 7. 10. 1L 12. 13. R 15. 16. 17. 18. 20. Dcscriptiou: The graph is the segment whose endpotnts are A{2, 3) and B(0, 3). x > -2 and y = 2 x = 5 and i/ > *>3 K = -3 and -2 < r/ < 5 —5 <x< — lor2<i/<5 — 5 < x < — 1 and 2 < y < 5 Which, if any, of the following points ure not part of the graph for Ex- ercise 12: ( — 5. 2), ( — 1, 2), ( — 1, 5), ( — 5, 5)? Which of these points are not part of the graph for Exercise 13? If P = (2. 0) and Q = (9. 0), what is the length of segment ¥Q? Justify your conclusion. If A = (2, 5) and B a (8, 5), what is the length of ATJ? Why? If C = { -3, - 4) and D = ( -3, 6). what is the length of CD? Why? Give the ctHirdinates of the midpoints of the segments whose endpoints are the following: (a) (3, 2), (3. 12) (b)(M>,(9.4) (c) (-3,1), (7,1) chau-kn'CE probi-em. Give the coordinates of the midpoint of the segment whose endpoints are (3 T 5) and (8, 7). challenge pkoblem. If F = (5, 1) and Q = (8, 6), what is the exact length of FQ? 452 Coordinates m a Plane Chapter 1 1 11.4 DISTANCE FORMULAS Consider an ^-coordinate system as shown in Figure 1 J -12. Let J be the unit point on the it-axis and let P and Q be any two points with al jseissas % l and x 2 , respectively, on the .t-axis. '*>y i £ 9 x «1 3 x 2 r^giire IMS We know, by the definition of a coordinate system on OX relative to unit segment Ol (Definition 3.3), that PQ (in 01 unite) = s, - x 2 \ = (** - «,|. For example, if P = (3, 0) and Q = {7, 0% then %i = 3, x* = 7, and PQ = |3 - 7| = |7 - 3J = 4. Now suppose that I is any line in the xy-planc and parallel to the .T-ajds as in Figure 11-13 (that is, I is any horizontal line), Let P and Q be any two points on I and let Pi(x u 0) and Qi{x 2 , 0) be the projections of P and Q t respectively, on the x-axis. Q 3— t* f, 0: ■i": i 2 Figure 11-13 By our definition of the abscissa of a point in an ry-plane, we know that the abscissa of Pisxj and that the abscissa of Q is x 2t If P = Pi and Q = Q u then PQ = PyQi = fa - X2J. 11,4 W>tar»ce Formula* 453 If P=£P lf then Q=£Qi and the quadrilateral PPiQi() is a parallelo- gram. Therefore we again have PQ = P t Q, = \x, - x 2 \. Note that iflisa horizontal line, then / is perpendicular to the y-axis (Why?) and every point of I projects onto the same point in the i/-axis. Thus all points on a horizontal line have the same ordinate. We have proved the following theorem. TJJEOBE \f 11.2 If p( Xit y T ) and Q(x 2 * ij\) are points on the same horizontal line in an xi^plane, men PQ = |*i - *!. It should he noted that if P = Q in Theorem 11.2. then x, a Xt and PQ = = 0. TllEOR EM 11.3 UP(x l ,y 1 ) and Q(x if f/ 2 ) are points on the same vertical line in an .viy-plaiie, then pq - m - y«l- Proof: The proof is similar to the one given for Theorem 11.2 and is assigned as an exercise. Example I If A = (3, 5) and B = (-6, 5), find AB. Solution! AB = [3 - (-6)| = 9, Note that if two points have the same ordinate, then they lie on the same horizontal line. Thus A 6 in Example 1 is a horizontal line. Hence AB is die absolute value of the difference of the abscissas of the points A and B t as shown in the solution to Example 1. Example 2 If C = (-2, -3} and D = (-2, 4), find CD. Solution: CD m -3 -4 = |-7| = 7. Note that if two points have the same abscissa, then they lie on the same vertical line. Thus CD in Example 2 is a vertical line. Hence CD is the absolute value of the difference of the ordf nates of the points C and D, as shown in the solution to Example 2. 4W Coordinates in a PJarre Chapter 1 1 You have seen how to find the distance between any two points in an ay-coordinate system if the two points lie on the same horizontal line or if the two points lie on the same vertical line. Next we show how to find the distance between two points if Lhey lie on the same oblique line, that is, a line that is neither horizontal nor vertical Before pro- ceeding to the general case, let us consider an example. Exampk3 UP = (-2, -3} and Q = (4, 5), find FQ. Solution: {See Figure ! t - 14.) Let h be the line through Q and parallel to the y-axis and let h be the line through F and parallel to the x-axis. These twojines intersect in a point R such that h ± h at R, Why? Therefore FQ is the hypotenuse of a right triangle, &PQR, with the right angle at R, Since f 2 is a horizontal line and l± is a vertical line, we have PR = |4 - {-2}| = 8 QR = |5 - (-3)1 = 8 by Theorems 11.2 and 11.3. By the Pythagorean Theorem, to 2 = m? + (<?w, = 62 + 82, = 36 + 64, = 100, and hence PQ = 10, ,55 t*i i A mm / i | / / ' ' \ -A ! * i 4- ~H _ ld__ k -2,-3) ]R<A,-3) ~^ 1 1 I i Figure 11-14 Note that in finding PQ in Example 3 we made reference to a right triangle. We proceed now to the theorem that, will enable us to find the distance between any two points in any ^-coordinate system without reference to a right triangle. The formula in this theorem is often re- ferred to as the Distance Formula for point 1 ; in an xu-plane. THEOREM HA If F, = (*,, pj and P 2 = (x 2t y 2 ) are any two points in an xy-plane s then PiPi = V(*T- *a) a + (m - y 2 ) 2 . Proof: There are four cases to consider. Cam L Fi = F 2 . Case 2. F L and F 3 are distinct points on a horizontal line. Case 3. Fi and P 2 are distinct points on a vertical line. Case 4. Pi and F 2 are distinct points on an oblique line. 11,4 Distance Formuttt 455 Proof of Case 1: If Pi = P** then x± = xj, yi = yz y and, by the for- mula of Theorem 11.4, we have P& = Vo = o, as it should since the distance between a point and itself is defined to be zero. Proof of Case 2: If P t and ?2 axe distinct points on the same hori- zontal line, dien y± = tj2 and, by the formula of Theorem 1 1 .4, we have PlP2 = \Z(X! - X 2 )2 = h - X 2 . This result agrees with the statement of Theorem 11.2 for two points on the same horizontal line. Proof of Case 3: If Pi and P 2 are distinct points on the same vertical line, then Xi = x% and, by the formula of Theorem 11.4, we have PiC«i.ja P1P2 = y/(y L - 1/2) 2 = |yi - U2I which agrees with the statement of Theorem 11,3. Proof of Case 4: If P t and P 2 are distinct points on an oblique line as shown in Figure 11-15, then the line through Pi and parallel to the y-2Lsis intersects the line through P% and parallel to the x-axis in a point H(xy, r/2) such that APiP^H is a right triangle, P 2 R = [xi - %z\ by Theorem 112 and Piil = Ij/j. - y a \ by Theorem 1 1 .3. We have (p 2 Rr- = ,*, - x 2 p = (x, - and Since PiPz is the hypotenuse of AP1P2R, we have, by the Pytha- gorean Theorem, that J^*2.j'a) or that PxP* ~ V and the proof is complete. (P1P2) 2 = (P 2 H) 2 + (PxH) £ fi - X2P + (#, - ya) 5 . 456 Coordinates in a Plane Chapter 11 Since (xi - $$* = (x 2 — *i) 2 and (y t - y 2 ) 2 m (y 2 - ifi) 2 . in using the Distance Formula of Theorem 11.4, it does not matter which point is designated P t (x lt tji) and which point is designated Pg(* 2 » y 2 )> Example -1 If A = (2, -4) and B = (-5, 3), find AB> Solution: Substituting the coordinates of the given points in the Dis- tance Formula, we have AB=W- (S)p + (-4 - 3)4 = yya + (_ 7) 2 = y/49 + 49 1 = 7\/2 Note that in working Example 4 we considered A(2, — 4) M the point J*i(xi, 1/1) and B( — 5, 3) as the point Pafet Jte) when we substi- tuted these coordinates into the distance formula. Show, hv working Example 4 again, that we would have obtained the same result for AB had we considered A(2, — 4) as the point Pz(x 2 , y 2 ) and B( — 5, 3) as the point Pxfa, ^i). EXERCISES 11.4 1. Prove Theorem 1].3. In Exercises 2-10, use Theorem 1 1 .2, Theorem 1 1 ,3, or the distance formula to find the distance between the given points. 2. (-3, 2) and (-3, 11) 3. (f I© and <& ID 4. (-2.5, V^ and (17.3, y5) 3. [w, 4.8} and fa -9.6) 6. (-2, 7) and (3, -5) 7. (3. -6) and (9, 0) 8. (4 r 17) and (-3,9) 9. (-3, -5) and (5, -1) 10, (6, -3) and (-4, 2) IL Find the perimeter of die triangle whose vertices arc A( — 2, —3), B(3, 9), and C(- 10,12). 1L4 Distance Formula* 457 12. Prove that the triangle whose vertices Are P{1> 2), Q(9 t 2), and /l(5, 8) is isosceles. 13. AAHC lias vertices A (6. 0), B(-4 t 4), and C(10, 4). (a) Find the perimeter of A ABC. (b) Find the area of AABC. 14. Find the lengths of the diagonals of a quadrilateral ABCD if A = (4, -3), B = (7, 10), C = (-8, 2), and D = (-1, -5}. 15. The vertices of APQR are P(- 1, -2), ()(4, 0), and fl(2, 5). Prove that A PQR is a right isosceles triangle. 16* If the distance between A(B t -2) and B{0 t tj) is 10, find the possible tf^oordtnates of B* 17* Find the coordinates of the points on the x-axis whose distance from (2, 8) is 10. IS. If (a, — a) is a point in quadrant IV, prove that the triangle with vertices (—5. 0) f (0, 5), and (o T — a) is isosceles. 19. Given D = (-2, 2), E = (10, 2}, F = (4, y) with Z DKE a right angle, find the two distinct possible values of jf. 20. Given P = (-2, -7), Q = (3, 3), R = (6, 9), use the distance formula to show that PQ + QR = PR, and hence that F-Q-R. ■ Exercises 21-26 refer to the triangle whose vertices are A = (2, 4), B = (6, 8), and C = (12, 2). 21. Draw line /i through A and parallel to the r/-axis. Draw line / 2 through B, parallel to the x-axis, and intersecting f t at D. What are the coordi- nates of D? 22. Draw line h through C, parallel to the x-axis, and intersecting h at E. What arc the coordinates of £? 23. What kind of quadrilateral is quadrilateral BDEC? Find \BDEC |. 24. What kind of triangles are ABDA and ACEA? Find \ABDA\ and AGFA . 25. Copy and complete: BDEQ m \AABC\ + \A[T\\ + \&JB , 26. Find \AABC\. 27. challenge problem. Find \APQR if P = (6, 0), Q = (1, 5), and R = (10, 8). 2& CHAU.EXCE problem. If A = (0, 0), D = (b t c), B = {«, 0), and C = (a ■+■ b* c), where a, b t c are positive numbers, prove that quadrilateral ABCD is a parallelogram. (Hint If the opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.) 4M Coordinates in a Plane Chapter 11 11.5 THE MIDPOINT FORMULA We know that the midpoint M of a segment PQ is the point be- tween P and Q such that PM = MQ or thai PM = JPQ, or, similarly, M@ = ^P(X Suppose that the coordinates of P and Q arc given and that we wish to find the coordinates of M, the midpoint of J 5 ^. For ex- ample, if P = (2, 3) and Q = (8, 5), we can find the coordinates (# t/) of M, the midpoint of FQ t as follows. (See Figure 11-16.) '■' 1 aotn t M"i0.y) Mix: ■-<* £r l«*3> 1 ! i i i X ^(2,0) M'{x,0) Q'£8,0) | 1 ! ! Ffcm Hie Let F(2 T 0), AT(x, 0) t (X(8, 0) be the projections of P, M, Q, re- spectively, on the *-axis and F'(0, 3 )* W'{0, i/), Q"(0, 5) be the projec- tions of P, M, Q f respectively, on the y-axis. Since If is the midpoint of FQ, it follows from Theore m i 10. S that M' is the midpoint of FQ' and M" is the midpoint of Fp 77 . Therefore, by Definition 3,3, 2 < x < 8 and 3 < t/ < 5< By the definition of midpoint, we have (1) FW = M'Q' and (2) V'M" s M'V ■ By Theorem 11.2, FM' = \x - 2| and Afp* = |8 - *|- Sinee x — 2 > (Why?) and 8 - a: > 0, we get by substitution into (1), x - 2 = 8 - x. Therefore 2x = 10 and x = 5. Similarly, by Theorem 11.3, F'M" = \y-3\ and M"Q' f = \5 - y\. Since y — 3 > and 5 — y > 0, we get by substitution into (2), y - 3 = 5 - y t Therefore 2y = 8 and y = 4 Hence the coordinates of .W, the mid- point of PQ, are (5, 4), 11.5 Th« Midpoint Formula 459 We can proceed, as in the preceding example, to find the coordi- nates of the midpoint of any segment if the coordinates of the end- points of the segment arcs given. However, Theorem .1 1.5 provides us with a formula that will enable us to find the coordinates of the mid- point of a segment. The formula in Theorem 1 1.5 is often referred to as file midpoint formula. THFnREM 1L5 UP = {x ly yi) and Q = {x 2 , y 2 ) are any two dis- tinct points in an xy -plane, then the midpoint M of FQ is the point »= fr^> There are three cases to consider. Cose 1: P and Q are distinct points on a horizontal line. Case 2: P and Q are distinct points on a vertical line. Case 3: P and Q are distinct points on an oblique line. We shall begin the proof for Case 3 and assign the remainder and the proofs of Cases 1 and 2 as exercises, Procifof Case 3: Let P(xi,y t ) and ()(X2, 1/2) be any two distinct points on an oblique line in an sy-plane as shown in Figure 11-17 and let M(x, y) be the midpoint of FQ. Let F(xi, 0), M'(x, 0), and Q'{x 2 , ()) be the projections of P> M, an d Q, r espectively, on the %-axis. By Theorem 10.3, M' is the midpoint of r<^ r . By Definition 3.3, x is between xi and X% dius <*< x% or X'l < * < Xt< (In Figure 11 -17 we have shown x± < x < x 2 , but this order might be reversed if P and Q are chosen in a different way.) <jfa2,^) F(*1,Q) M'lx,0) Q-Cx 2l CI) Figure IM7 460 Coordinates in a Plane Chapter II By the definition of midpoint, par = arg\ and by Theorem 115, FAT = \x - %i\ and M'Q f = ,** - 4 Therefore \X - Xi\ = |-t2 - X\> If *i < i < x 2 , then x — xi > and *a — x > and we have x — *i = x 2 — x. Why? Therefore 2x = *t 4- x 3 and " = —2— If ^2 < x < xi, then xi — x > and x — x 2 > 0. In this case, FM' = |x - ri. = xi — x MV = |x 2 - x\ = x - z 2 . x — x 2 = Xi — x, and again we have -_ * + » 2 ' We have shown that if M (x, y) is the midpoint of the segment whose endpoints are P{x yy iji) and Q{x 2 , ys), then the abscissa of \f is * 2 . In a similar way, it can be shown that the ordinate of M is ** 1 _ ^ 2 , (Yon arc asked to show this in the Exercises, thus completing the proof of Case 3.) hxamplc 1 Find the coordinates of the midpoint of the segment whose endpoints are the following: 1. A(2. -3) and 5(2, 9). 2. C(-12, 1) and D(-3, 1). 3. E(-2, 7) and F{W t 12), 11,5 Th* Midpoint Formula 461 Solution: L Segment XB lies on a vertical line. Therefore i = x\ = *t and *i + *£ *i + JCi —IT = ~2— "* Therefore the midpoint is 2. Segment CD lies on a horizontal line. Tlierefore y = t/i = 1/2 tfi + ya _ yi ■+■ tji __ 2 "-* 1 ' 2 Therefore th« j midpoint is Ma V 2 * yi) /-12 + (- -3) .»)-(-*»> 3. Segment EPlies on an oblique line; hence the midpoint M of EF is the point Example 2 The vertices of a triangle are A(0, 5), B(4 f 3), and Q-2, 1). Find the length of the median to BC. Solution; The median to BU is the segment whose endpoints are A{Q t 5) and M, the midpoint of BC, The coordinates of M are (1, 2) by the midpoint formula; hence AM = y/(Q - If + f£ - 2) 6 = VTT9 = V10 by the Distance Formula. 462 Coordinates in a Plane Chapter 1 1 EXERCISES 11.5 1 . Complete the proof of Case 3 of Theorem 11.5 by showing that the ordi- nate of M is Hl±Jil , (See Figure 11-17.) IS 2. Prove Case I of Theorem 11.5. (In this case, y = tj\ = tj2 in the state- ment of the theorem.) 3» Prove Case 2 of Theorem 1 1 .5. (In this case, x = i] = r* in the state- ment of the theorem.) ■ In Exercises 4-10, find the midpoint of AB if A and B have the given coordinates, 4. (-5, -2) and (-5, 6) 5. (-3, 5) and (8,5) 6. (0, 0) and (8, 10} 7. (0,0) and (-8, 10) 8. {1, 2) and (6, 14) 9. (r, 7) and (3r, -3) 10. fab)md(-5a,7b) 11. The vertices of a triangle are I\2, -3), £{10, 1), and K(4 t 6). Find the midpoint of FQ. Find the length of the median to FQ. 12. The vertices of a triangle are A(-3, -2), B(l, 6), and C(5, -2), Find the lengths of the medians lo A~B and BC« How are the lengths of these two medians related to each other? What kind of triangle is A ABC? ■ Exercises 13-15 refer to quadrilateral ABCD whose vertices are the points A = (0, 0), B = (6, 0), C = {8, 4), D = (2, 4). 13. Find AC and BIX 14. Show that the midpoint of AC is the same point as the midpoint of B7X 15. What kind of quadrilateral is ABCD? M E*ercises 16-24 refer to A ABC whose vertices are the points A = (2, 0), B = (12, 0), C = (7, 5\/3). I K. Prove that A ABC is equilateral. 17. Find the coordinates of D, the midpoint of AB. 18. Find the coordinates of E, the midpoint of fflOL 19. Find the coordinates of F, the midpoint of AC. 20. Show that the lengths oi' the three medians of A ABC are equal. 21. Find DE and show that t>E = JAC. 11.5 The Midpoint Formula 463 22. Find EF and show that EF = JAB. 23. Find FD and show that FD = \BC. 24. Do the results of Exercises 21-23 prove that ADEF is equilateral? Exercises 25-27 refer to A/SK whose vertices are the points / = (0, 0\ S = (6, 0), K = (0 25. What kind of triangle is A JSK? 26. Find the coordinates of M, the midp>int of 5K. 27. Prove that JM s ]$K, In Exercises 28-31, the coordinates of two points A, M are given. Find the coordinates of the point B suck that M is the midpoint of 3uB, 28. A = (1,3),M={4,7) 29. A = (4, 7), Af = (L 3) 30. A = (-1, -8), M = (0, 0) „ 31. A = (-6, -4), M = (-3, -2} Exercises 32-36 refer to the segment AH whose endpoints arc A = (3, 2), B= (11, 6) and the point Pi x> y) on AB such that AP = %AB. 32. Let A', P\ /T be the projections of A, J* B. respectively, on the .t-axis. 33. What is the absciss u of F in Exercise 32? What is the abscissa of P? 34. Let A", P", B" be the projections of A , P, B t respectively, on the (/-axis. 35. What is the ordinate of P" in Exercise 34? What is the ordinate of P? Mi, What are the coordinated of PI 37. If A = (-4, -2) and B = (6, 3), find the coordinates of the point P on AB such that AF = JAR (ERnfc See Exercises 32-36.) 38. challenge problem. Given positive numbers a and h and a right tri- angle with vertices at A = (0, b) t B s (a, 0), and C = (0, 0), find the coordinates of Af, the midpoint of A~B t and show that CM = ^AB. Does tills prove that for antj right triangle, the median to the hypotenuse is one-half as long as the hypotenuse? 39. challenge PROBLEM. Given positive numbers a, h, and u quadri- lateral ABCD with vertices at A = (0, 0), B = (a, 0), C = (a + b, c), and D = (b, c), (a) prove that ABCD is a parallelogram, (b) prove that the diago nals AlT and ED bisect each other by showing that the midpoint of AC is the same point as the midpoint of B~D. 464 Coordinates in a Plane Chapter 11 11.6 PARAMETRIC LINEAR EQUATIONS If we are given a line in an xjy-ptune, it is often desirable to find the coordinates of points on that line. We know that two distinct points determine a line. Suppose that we know the xi/-coordinates of two dis- tinct points A and B on AB. We should be able to find the xiy-coordi- nates of any other point P on AB provided, of eoursc 5 enough informa- tion is given to determine one and only one point P on AB. Example I Let A = (1, 3), B = (4, 7), and suppose that S, Q, fl are ^ -y v ^ fr points on AB such that 5 is on AB, Q is on AB, and R is on opp AB, [See Figure 11-18.) Suppose, further, that AS = $AB, AQ = 2AB, AB. = AB, and we wish to find the coordinates of S, Q t and iL 4 iy J* J 7$ / . / 1 1 *£:- 6 Ml X s ;i / B) flfa X ~ 3 di b / ~ 3 Figure 11-18 Let S = (x, y). To find x and y we might proceed as follows. Let A\ B', S f be the projections of A, B % S, respectively, on the r-axis, and let A", B" t S" be the projections of A, B, S, respectively, on the y-axis. Using Theorem 10.8, we get 2 _ 3 " AS _ A'S' *— 1 AB " A'B' " 3 2 3 " AS A"$" y - 3 AB A"B" 4 and and .r = 3, 9 3 11.6 Parametric Linear Equations 465 Similarly, we can find that Q = (7, 11) and R - (-2, - 1). (Verify these results by selling up appropriate equations and solving Ihcm.) If Fwere any other point on AB such thai AP=kAB, where k is any positive number or zero, we could find the coordinates of P by a computation similar to those above. Our objective, however, is to derive an expression from which the coordinates of any point on AB can be obtained by simple replacements. In Chapter 3 we studied a coordinate system on a line. In this chap- ter we have already defined an xi/-coordinate system in a plane, based on two line coordinate systems like those you studied in Chapter 3, Let us consider now a third line coordinate system, a coordinate « — * system on line AB with A as origin and B as unit point We call it the ^-coordinate system on AB, It should be clear that we start with an ,tf/-coordinate system based on a unit segment for distance. When we speak of "the distance*' between two points in die .ry-plane, we are talking about the distance based on the same unit segment that is used in setting up the x^-coordinate system. For every different choice of points A and Bona line fin the i^-planc, there is a different k coordi- nate system on I with A as origin and B us unit point. The ^-distance between two points on / will usually be different from the ^(/-distance between those points. The following table shows the ^-coordinate, the {/-coordinate, and the ^-coordinate of the points A, B, S, Q, B that were shown in Exam- ple L Point x-Coordinate y-Coordinate fe-Coordiiiatc A 1 3 B 4 7 1 S 3 s§ I Q 7 11 2 R -2 -1 -1 P m 13 k We shall derive the equations which show us how to compute the x- and {/-coordinates of a point on AB in terms of its fc- coordinate. First we shall show that the ^-coordinates of the points of I form a coordinate system on I as do the ^-coordinates. .:i,i. Coordinate* in a Plane Chapter 11 Let / be any nonvertical line in an .vy-plaue. (See Figure 11-19.) Let O(0, y ), fft yi), P(x, y) be points on l y and let O t l\ F be their re- speetive projections on the x~axis. It foflows from the Plane Separation Postulate and the properties of parallel lines that (J r t t F have the same betweenness relation as do their respective projections. For ex- ample, if O-I-F. then O-l'-F. Figure IMS Let § denote the unique coordinate system on I with O as origin and I as unit point. Note that Ul serves as a unit segment for § and that generally (except if / is parallel to the Y-axis) distances "in g" arc dif- ferent from distances "in the xtf-coordinate system.*' Let p be the co- ordinate of F in the system §, Since betweenness foT points on I is the same as for their projections on the x-axis, it follows from the definition of a coordinate system on a line that p and x are both positive, or both negative, or both zero- It follows from Theorem 10.8 that OP O'F or ' Since these ratios are equal regardless of the distance function, we have OP OI ;i _ 1 ir " or "1-0 = Jx|, and p as x. Now S is the one-to-one correspondence that matches each point P on / with a number p. Since p = x, we see that the correspondence that matches each point of I with its x-coordinate is indeed a coordinate system. Similady if lis a non horizontal line, then the correspondence that matches each point of I with its y-ccKM-dinate is a coordinate system on I. We state these results in our next theorern. TlfEQREM 11.6 If I is any nonvertical (nonhori/ontal) line in an xiy-planc, then the one-to-one correspondence between the points of / and their x-coordinates ((/-coordinates) is a coordinate system on I. 11.6 Parametric Linear Equations 467 In the next theorem, as well as in many others throughout the rest of this book, there is the assertion, 'A is real" within a set-builder sym- bol. Ttiis is short for "k is a real number/' THEOREM IL7 If A(x ls yi ) and B(x 2t y 2 ) are any two distinct points, then AB = {(x y) : x = x, + % 2 - XjJ.y = {/! + %a - J/i),fcisreal}, If fc is a real number and if P is the point (x t y) where x = *i + fc(x 2 - «!) and ^ = yi + k(y 2 - fiX then fe a AT and P€ A§ if fc> 0; "*-4l and P£oppA$ if Ar < 0. JVoo/: Let Aixi, y t ) t B{x 2 , f&) be any two distinct points. Suppose first that AB is neither vertical nor horizontal, Think of three coordinate systems on AB as suggested in Figure 1 1-20, the x-eoordinate system and the y-coordinate system, determined by the xy-coordinatc system (see Theorem 11.6), and the ^-coordinate system in which A is the origin and B is the unit point. Figure 11*20 or or 468 Coordinates in a Plane Chopttr 11 It follows from tlic Two-Coordinate-Systems Theorem (Theorem 3.6) and its corollary, each applied twice, that AB is the set of all points P(x 3 r/) such that x — xi k — y — iji _ k — x-2 — xx 1 — * y2 — yi 1 — * x - xi = *(* 2 - Xi\ y - i/! = % 2 - yt); x = X! + *<*2 - *i). y = yi + %2 - yi); where ^j^. Since the definition of a coordinate system requires that hetween- ness for points agree with betweenness for coordinates, it follows that P is on a5 if and only if h ;> 0, and ? is on opp AB if and only if k < 0, If A E is a vertical line, then the x-coordinate of every point P on AB is die same number, and there is no x-eoordinate system on Ad, In this case xi = xo = x for every point P{x t y) on AB and the equation x = *! + *(* 2 - *i)* which simplifies to i = ii, is still applicable. The proof for the noii- vertieal-nonhorizontal case is applicable to this case as far as the rela- tion between y and k is concerned. Therefore the assertion of the the- orem applies in the vertical case. Similarly, the theorem may be proved for the horizontal case. Tn S3 = {(x, y) : x = x t + k(x 2 - %§}, y = y± + k(y 2 - yi), k is real}, the "it is real" is put in because k is not mentioned before the braces and not before the colon within the braces, and we read the sentence as "AB is the set of all points (x t y) such that x = X! + k(x 2 - x a ), y = yi + %2 - yi), where k is real/' The equations X = x l + k(x-i - Xi) and y = i/i + %z - f/i) hi the statement of Theorem 11.7 are called parametric equations for die line AB and k is called the parameter. A parameter j s usually 11.6 Parametric LI near Equations 469 thought of as a variable to which values may be assigned arbitrarily, and other variables are defined in terms of it. By assigning real number values to the parameter k in the parametric equations for AT?, we ob- tain ordered pairs of numbers corresponding to points on AB in the xy- eoordinate system. Of course, we could assign values to x (in the non- vcrtieal case) and find values of y so that the resulting ordered pairs correspond to points of AB or we could assign values to y (in the non- horizontal case) and End values of x so that the resulting ordered pairs correspond to points of AB. But it is easier, and it works in all cases, to assign values to k and compute values of x and of y so dial the resulting ordered pairs correspond to points of AB. Note that parametric equations for a line aie not unique. If / is a line, then there are many choices for the two points A and B of The- orem 11.7, and hence many pairs of parametric equations for L If we replace the coordinates (% x 3 ) and (yj, j/g) in the parametric equations of Theorem 1 1 .7 with the coordinates (1, 3) and (4, 7) of the points A and B in Example 1, we obtain the parametric equations x = 1 + 3k and y = 3 -I- 4k for the line AB in that example. Recall that, in Example 1, S is a point on AB such that AS = |AB. Thus, by replacing k with the number f In the parametric equations X m 1 + m, y = 3 + 4Jt, we obtain the coordinates (3, 5j) of the point S. Show that you get the xy-coordinates of the points Q and R on X5 in Example 1 by assigning the values 2 and — 1, respectively, to k in the parametric equations x = 1 + 3fc, y sb 3 + 4k. It follows from Theorem 1 1,7 that every line can be represented by set-builder notation and a pair of parametric equations. However, it is not true that every pair of paTametric equations represents a line. For example, the set S = {(*, y) : x = 2 4- k ■ 0, y = 3 + k ■ 0, k is real} is a set whose only clement is (2, 3). Note that in the statement of The- orem 11.7, the points A and B are distinct; hence not both of the co- efficients of k in the parametric equations can be zero. That is, we can have x 2 — xi=0 or y2 — tji=0, but we cannot have both x$— xj = Q and y2 — yi = 0. 470 Coordinates In a Plane Chapter 1 1 Although not eveiy pair of parametric equations represents a line , our next theorem provides us with a method for identifying those pairs of parametric equations that represent lines. THEOREM 11.8 If a t h, c, d are real numbers, if b and d are not both zero, and if S = {(*, y) : x = n + M, y = c + dk, k is real}, then S is a line. Proof; Taking k = and k = 1, vvc get two points in S, namely A (a, c) and B(a + b t c + rf )• Tty Theorem 11.7, parametric equations for AB are sc = fl-r*ft(a-Hfc — a) = a + M and tfs«-rx(6+a-c)s@-fJ& Therefore A3 = {(x f y) ; x = a + bkj y = c -\- dk,kis real}; hence S = An and S is a line. hi addition to being able to write parametric equations for a line, we can also write parametric equations for a segment or a ray if we place the proper restrictions on the parameter k. We illustrate this technique in the following examples. Example 2 Let A = (4, 1) and B = (2, —3). Using coordinates and parametric equations, express (1) AB, (2) AB. and (3) AB. Solution; 1. Substituting the coordinates of A and B into the equations of Theorem 11,7, we get x = 4 - 2k and y = 1 - 4k as para- metric equations for AB. Therefore AB ' = {(x, y) : x = 4 - 2k, y = I - 4k t k is real}. 2. If P is a point on AJ?, then P £ A? if and only if Jt > 0, Therefore AB = {(*. y) : * = 4 - 2fe y = 1 - 4^ * > 0}, 3. If P, A, B are three points on AB and if &i, & 2 , *a are the coordi- nates of P, A, /I, respectively, then Pis between A and # if and only if fci is between feg and fife. Since the ^-coordinates of A and B on AB are and 1 , respectively, then P is between A and B if and only if k is between and 1. Therefore m = {(x t y) : x ^ 4 - 2k >y = I - 4^0 < k<, } }. 11.6 Parametric Lln«ar Equations 471 Example 3 Given A = (4, 1} and B = (2, -3), find P on AB such that AP = 4 'AB, Solution: Taking k = 4 in the parametric equations of Example 2, we get P = fry) = (-4, -15). Example 4 Given A = (4, 1 ) and B = (2 t - 3), find P on opp AB such AP = 4 • AB, Solution: Taking k = — 4 in the parametric equations of Example 2, we get P = fr y) = (12, 17). Example 5 Given A = (4, I) and B = (2, -3), find C and D % the points of bisection of AB. Solution: Taking fe = -j- and ft = -J in die parametric equations of Example 2, we get C = (J£ , — -§-) and D a (|, -}) as the points of bisection. EXERCISES 1LG In Exercises 1-5, the coordinates of two points A, B are given. Use para- metric equations and set-buiJdcr notation to express AB t AB, opp AH, and m L A = (1, 4), B = (3, 7) 4, A = (2, 3), B = (0, 0) 2. A = (2. 2). B = (5, 5) 5. A = (-3, -2), 3 = (0, 1) 3. A =(-1,3), B = (3,0) G-IO* Find the coordinates of the midpoint of AB in Exercises 1-5. {Hint: Let k = -j in the parametric equations for AB.) 1 1. Find the points of bisection of AB in Exercise 2. hi Exercises 12-17, a relation between AP and AB is given, If A = (-2, -5) and B = (4. 1), find the coordinates of P in Z8. 12* AP = 2AB 15, AP = t^2AB 13. AP = 25AB 10. AP = LSAff 14, 4AP = MB 17. AP = \AB 472 Coordinates in a Plane Chapter 1 1 18-23. The instructions for Exercises 18-23 arc the same as for Exercises 12-17, except that F is in opp AB, Kecall that -j^ = —kin this case. In Exercise 18. AP = 2AB as in Exercise 12; in Exercise 19, AP = 25AB as in Exercise 13; etc. 24. Find the coordinates of P £ AB if A = (-1. 5), B = (4, -3), and AP = 3A#. (There are two possible answers.) 25. Given A = (0, 4), B = (3„ 0), and (7 is on AB, find the ^-coordinate of C if its x-coordinate is — 2. (Hint: Obtain the parametric equations for AB and let x = —2 in one of these equations (Which one?). Solve this equation for k and use this value of k to find the (/-coordinate of C.) 2a Given A = ( - 1, 3), B = (2, -3). and C is on A$ t (a) find the {/-coordinate of C if its x-coordinatc is 5. (b) find the abscissa of C if its ordinate is 8. In Exercises 27-31, draw the graph of the set. 27. {(*, ij)\\+3k,y = 2- k, k is real} 28. {(x, y}:x = 3k.y = k,0<k<3} 29. {(x, y):x= -2 + k,y = -2k, k > 0} 30. {(*, y) : x = -K y = K * <, 0} 31. {(x, y) : x = 5, y = 2 + Jfc, -2 < k < 3) Exercises 32-39 refer to the triangle whose vertices are A = (3, 2). B = (9, 4), and C = (5, o), 32. Kind the coordinates of D, the midpoint of AB. 33. Find the coordinates of E, the midpoint of BC, 34. Find the coordinates of t\ the midpoint of AT. 35. Find the coordinates of the point R in CzJauch that CR s 5CD. (Use the parametric equations for CIJ, where C =■ (xi, yi) = (5, 8). D = (x 2 , i/ 2 ) = (6, 3), and A- = f ) 36. Find the coordinates of the point S in BF such that #S = JBF. (Use the parametric equalions for BF, where B = (xt, t/i), F = (x 2 , 1/2), and 37. Find the coordinates of the point Tin AE such that AT = JAE. 38. Is Jt = S = 77 39. Show Uiat tlie three medians of A ABC intersect in a point whose dis- tance from each vertex is two-thirds of the length of the median from that vertex. 40. challenge problem. If a, b, e are positive numbers and if A = (0, 0), B == (a, 0),. and C = (b, c), show that the three medians of A ABC are concurrent at a point whose distance from each vertex is two- thirds of the length of the median from that vertex. 11.7 Slope 473 11.7 SLOPE In this section we develop the idea of the slope of a line. Slope corresponds to the idea of the steepness of an inclined plane or the steepness of a stairway. If all the steps of a stairway are uniform, we may describe the steepness of the stairway in terms of the "rise" and "run** of one of its steps. The steepness of a uniform stairway is the number obtained by dividing the rise by the run of one of the steps. For example, we may say that the steepness of the stairway shown in Figure 11-21 is |. Figure U-Sl In our formal geometry, we define the steepness or slope of a line in a similar way. That is, we define the slope of a line in terms of the slope of any segment on the line and we define the slope of a segment in an xy-plane in terms of the coordinates of its endpointa. In informal geometry, we can think of the slope of a segment in somewhat the same way as we think of the slope of a step in a stairway; that is, in terms uf its "rise divided hymn." Figure I 1-22 suggests that the slope of AJ9 is $ = -J-. Note that in terms of the coordinates (2, 3) and (8, 7) of the endpoints A and B of AB, the rise is |7 — 3| = 4 and the run is |8 - 2| = a Figure 1 1-22 kj - 1 1" V*M H Ri j* 90 I 3 .- 1 . A(*3) Run C{8, 3) v k 9 Coordinates in a Plane Thus, if A = (x±, 1/1) and B the slope m of A3 as Chapter 11 = (x^y j/y), where *i =?^ *2» we c.otdd define m = x a — Kl hut we do not. The same formula without the absolute value symbols is not only easier to handle but it is more useful Hie sign of the slope Indicates whether the segment "slopes up or down." In Figure 11-22, segment AB has a positive number for its slope and we note that the segment slopes up as we view it from left to Tight. However, the seg- ment CD shown in Figure 1 1-23 slopes down as we view it from left to right - '■• 111 I ioMB TCI — a 1 \ ! \ BWU \ D(7 r 1) J l B 1 | * ■ Figure 1143 If we use the formula x 2 - Xl in compute tlie slope of CD, we obtain 1-5 7-4 4 3' a negative number. This is one of the reasons why we do not include the absolute value symbols in our formula for the slope of a segment. That is, if the slope of a segment is a positive number, the segment slopes up when viewed from left to right, and if the slope is a negative number, the segment slopes down when viewed from left to right. We are ready now for our formal definition. Definition 11.2 If A(xi, y%) and B(% j/a) are two distinct points and if Xx =£ x*, then the slope of AB is * /g "~ * 1 . X-i — Xi If A(xu tfi) and B{x% t tjz) are two distinct points and X\ = #2, then "KB is a vertical segment and slope is not defined for 'KB in this case. 11.7 Siope 475 If A(*i, i/i) and B(x2, 1/2) are two distinct points and r/j = y 2 * then^B is a horizontal segment and its slope is zero. Note that in computing the slope of a segment A B it docs not mat- ter wiiich endpoint of AB is designated (afe 1/]) and which endpoint is designated fa, 1/2)- Thus, for the slope m of AB in Figure 11-22, we could write m = y*_zJ± = LzJL = 1 = 1 ' x 2 - xi 8 - 2 "" 6 "" 3 or we could write __ j/g - yi 3-7 _ _-4 _ 2^ X2 - Xi " 2 - 8 " -6 ~ 3 * Similarly, for the slope m of CD in Figure 1 1-23, wc could write Xz — Xx 7-4 3 or Xb — xi 4—7 3 As suggested earlier, the concept of the slope of a line is based on the concept of the slope of a segment. It seems reasonable that the slopes of all segments of a nonvertical line are equal, and we state this as our next theorem. THEOREM LL9 The slopes of all segments of a nonvertical line are equal. Proof: Suppose that AB is any nonvertical line with A = (x u y t ) and B = (x$, ys). Then, by Theorem 11.7, AB can be expressed parametri- cally as = {(x, y) : x = xi + k{x- 2 - x T )» ij s t^ -|- % 2 - ^i), fc is r By Definition 1 1 ,2, the slope of AB is X 2 — XL Let R and S be any two distinct points of AB. There are two distinct numbers k] and h$ such that R = (x-i + fei{xa - *i), i/! + Ai(y 2 - J/i)) and 476 Coordinates in a Plane Chapter 1 1 By Definition 11.5, the slope of ESh fa + k 2 (y 2 - iji)) - (y-i + Ky 2 - j/i)) _ (</a - t/i)fe - fr) (*"i + k 2 ix 2 — xij) — (x\ + k\(x2 — *i)) {x 2 - xt)(ki - ki) X2 - «J We have proved that any two -segments of AB have the same slope, and !hr pn lof is complete. Definition 11.3 The slope of a oonvcrtical line is equal to the dope of any of its segments. The slope of a non vertical rav is equal to the slope of the line that contains the ray. Example 1 Find the slope of the line which contains the points A = (-2,5) and B = (4,0). Solution: AB is a nonvertieaJ line and, by Definition 11.3, the slope of AB is equal to the slope of A~B. Therefore the slope of AH is y 2 - *Ji _ 0-5 _ _£ *2-*i "4-<-2)~ 6' We know that two distinct points determine a line in the sense that if any two distinct points are given, then there is exactly one line which contains them. It is also true that a non vertical line is determined by any point on it and its slope. That is, there is exactly one line which contains a given point and has a given slope. We now state tins formally, THEOREM 11.10 Given a point A and a real number m » there is one and only one line which contains A and has slope m. Existence. Let a point A = (*i, yi) and a slope m be given. Let B be the point (x\ + 1, «/i 4- m). Then AH is a line which contains A and has slope m. (Show that the slope of AB is m.) Uniqueness. Let PQ be any line which contains A and has slope m. Since FQ is a nonvertical line (Why?), it intersects the vertical line through B(xi + I, y 1 -f m) in some point R(xi — 1, y 2 ) as suggested in Figure 11-24. (The figure shows R and B as distinct points. We shall prove that they are actually the same point.) The slope of FQ is equal to 11.7 Slope 477 Figure U-S4 Therefore y 2 = jft + «b Hence H = B and Py sa AB, This proves that A/5 is the only line through A with slope m, and the proof is complete, If wc know the coordinates of two points on a line, we can use The- orem 11.7 to write parametric equations for the line. Theorem 11.10 implies that a line is determined by any point on it and its slope. There- fore we should be able to write parametric equations for a line passing through a given point and having a given slope. Our next theorem tells us how to do this. THEOREM I hi I The line I given by 1. I s {(*, y) : x = *i + fc, y = y, -f mk, k is real} or by 2. I = {{x, y) : x = xi + sk t y = y t + rfc, k is real) is the line through (x lt y\) with slope m = — . Fnx>f: We shall establish (J) and (2) separately. Troof of 1: Taking k = and k = I in the parametric equations in (1), we get (x% t i/i) and (*j -f 1, y± + m), two points on I. The slope of the segment joining these two points is yi + m - yi _rn_ m *i + 1 - xi 1 Therefore I as given in (1) is the line through (%, i/i) with slope m. 478 Coordinates in a Plane Chapter 11 Proof of 2: Taking k = and k a i in the parametric equations in (2), we get (*i, yi} and (x t + $, y^ + r), two points on I The slope of the segment joining these two points is j/i + r - yi _ r *i + & - Xy $ Therefore I as given in (2) is the line through {x^ t/i) with slope rn, and the proof is complete. Example 2 A line p passes through (2, 5) and has slope 3. Find the point on p whose abscissa is — 2. Solutimt: Using Theorem 11,11, we can express p as p = {(x, y) : x = 2 + ft, y = 5 + 3fc, k is real). We next set % = -2, obtaining -2 = 2-|-&,orfc= -4. Hence y = 5 + 3(-4) = -7. The point on p whose abscissa is —2 is (— 2, —7). Example 3 Find the slope of the line I = {(x, y) : x = 2 - 3fc # = 3, Jfc is real}. SoJi*rio»: Taking k = and k = 1, wc get (2, 3) and ( — 1, 3), two points on the line. The slope of the segment joining these two points is 3 - 3 _ -1-2 -3 Therefore the slope of f is zero. Is f a vertical, horizontal, or oblique line? We conclude this section with the following summary remarks re- garding the slope of a line. L If a line has a positive number for its slope, then the line is ob- lique and slopes up when viewed from left to right, (See Figure ll-25a.J 2. If a line has a negative number for its slope,, then the line is ob- lique, and slopes down when viewed from left to right, (See Fig- ure ll-25b.) 3. If a line has slope aero, then the line is horizontal. (See Figure ll-25c.) 4. Slope is not defined for a vertical line. 11.7 Slope 479 Zero ■lOM (C) Figure 11- EXKRCI5ES 11.7 In Exercises 1-10, find the slope of the segment joining the given |>oints. Express each answer as a fraction in lowest terms. 1. (2, 5) and (5, 7) & (2^ -4) and (-1, 4\) 2. {- 1. -3) and (8, 3) 7. (-5.6, 3) and (1.4, -£) 3. (0, -4) and (8, 0) 8. {1.2. -5) and (3.2, -5) 4. (- 1, 4) and (2, -2) 9. (2, 5) and (5, 2) 5. (-3, -3) and (3, 3) 10. (7. 12) and (1, -3) 11^20. In Exercises 11-20, tell without plotting the points whether the given line (a) slopes up, (b) slopes down, or (c) is horizontal. In Kxercisc 1 1, the line is the one that passes through the two given points in Ex- ercise 1. In Exercise 12, die line is the one that passes through the two given points in Exercise 2, and so on. 21. (a) On the same xt/-plane. graph the line r through the two points of Exercise 1 and the line s through the two points of Exercise 2, (b) What appears to be true about these two lines, that is, how are r and ^ related? (c) How are the slopes of these two lines related? 22. (a) On the same xy-plane, graph the line p through the two points of Exercise 3 and the line q through the two points of Exercise 4. (b) What appears to be true about these two lines, that is, how are p and q related? (c) How arc the slopes of these two lines related? [Hint: Find the prod- uct of the slope of p and the slope of q.) In Exercises 23-27, find x or y (whichever is not given) so that the line through the two points will have the given slope, 23. (4, 1) and (x, 4), m = 3 26. (4, -3) and (0, y), m =s -| 24. (-3, 1) and (6 t y) t m s -\ 27. (5, 12) and ( -2, y), m - 25. (x, 0) and (-3, 5), m = J 480 Coordinates in a Plana Chapter 11 28. (ti) On the same xtf-plane. graph the line I through the two points of Exercise 25 and the line n through the two points of Exercise 2ft (b) What appears to be true about these two lines, that is, how are J and n related? (c) How are the slopes of I aud n related? 29. (a) Plot the quadrilateral ABCD with vertices A = (— 2, —3), B = (3, -2), C = (4, 2), and D = (~1. 1). (b) Which pairs of sides have the same slope? (c) Is A BCD a parallelogram? 30. Write parametric equations for the line I through { — 2, 5) with slope 2. 31. Write parametric equations for the line p through (1, - 1) with slope -J. 32. Use the parametric equations for line / obtained in Exercise 30 to find the coordinates of at least one more point on I and plot the graph of 1. 33. Use the parametric equations for the line p obtained in Exercise 31 to find the coordinates of at least one more point on p and plot the graph of p. 34. One way to plot the graph of line p in Exercise 31 is suggested by the figure. Note thai the point Q{1, - 1) is on p and the "slope fraction" f tells us how to get from one point to another on p. The numerator 3 represents the difference in the ordinates of two points on p, and the de- nominator 2 represents the difference in the abscissas of the same two points. Thus, if we begin at the point (1, —1). which is known to be on p, we arrive at another point on p by increasing the ordinate and abscissa of [L — 1) by 3 and 2, respectively. If the slope is negative, it can be written as a fraction with a negative numerator aud a positive denominator; for example, —^—, In tliis example we could get from one point to another, in informal geometry, by moving 2 to the right and 3 down; in formal geometry, by adding 2 to the abscissa and subtracting 3 from the ordinate. Write the coordinates of the points A T B T C. . ":• fp % / / : ; i s 2 a 1 / J \ a / x % m- ■ 1) /I 11.8 Other Equations of Lint* 4-8 1 35* Use the method described m Exercise 34 to plot the graph of the line through (—2, 1) with slope y. 36, Use the method described in Exercise 34 to plot the graph of the line through (0, 6) with slope -J, 37, Use the method described in Exercise 34 to plot the graph of the line through (0. 0) with slope 2. (ffinf; 2 = f ) 11.8 OTHER EQUATIONS OF LINES In Section 11.6 we showed how a line can be expressed in terms of parametric equations. In this section we shall show how parametric equations for a line can lie used to obtain other equations of the line. You are already familial" with most of these equations from your work in algebra. Perhaps the simplest equations for lines are those for hori- zontal and vertical lines. THEOREM 1L12 If Ms the horizontal line through {x lf 1/1), then '= tfey) '■ y = »)■ Proof: Let (jcfc, 1/2) he any other point on the horizontal line I through the point [t\, iji). By Theorem 11.7, 1 = (fr y) : * = *i + H*2 - *i), y = yi + %s - yi)> k k real}. Tine I is horizontal so tj\ = y 2 - Therefore y = yi + %2 - #1) = !/i + *-0 = y T . For any real number fc» * s x* + k(x2 — x%) is a real number. Conversely, for every real number ,t» there is a real number k such that x = Xt + fc(*2 — x i)- Th^ one an ^ on ly ^ that works here is *2 - ^l It follows that I is the set of off points (at, «/) such that j/ = r/ t and x is a real number, that is, that This completes the proof. 462 Coordinates In a Plane Chapter 1 1 Iii the statement of Theorem 11.12 it should be clear that we could say that 1 = {(*> y) - x m Xi + k{x 2 - Xi), y = i fl , k is real}, or I = {(x t if) : x is real and ij = yi} t or ' = {(*; ST : y = §Ji. * is real}* Since the equation x = xi + A(* a — *i) establishes a onc-tc-otie cor- respondence between the set of all real numbers thought of as k- values and the set of all real numbers thought of as x-v&lucs. these set-builder symbols all denote the same set. THEORTM 1L13 If I is the vertical line through {x u ft), then / = {(x, y) ; x- *i}. Proof: Assigned as an exercise. Example 1 If P = { — 4, 3), write an equation of (1} the vertical line through P and (2) the horizontal line through P. Solution* 1. An equation of the vertical line through Pis x = —A. 2, An equation of the horizontal line through P is y = 3. Our next theorem tells us how to write an equation of a line if we know the coordinates of any two distinct points on the line. THEOREM 11.14 {The TwthPoint Form) Tf A = (x u y\) and B = (x2, 1J2) are distinct points, and if AB is an oblique line, then = {fr„) :£=*-= JLzIl) I X 2 — Xi 1/2 - Ui) *2 - x t y 2 - y^ Proof: We are given that A = {xi, tji) and B = (x z , ife) are distinct points on oblique line AB< There are two things to prove. 1. If P = (x, u) is a point of Aft, then — = " "~ - 1 - . 2 - If ^3^ = ^7=^ ^ P = ^ $ '^point of £2. UJB Oth« Equations of Lin** 433 Proof of 1; Suppose that F a (x, 1/) and that P is a point on A/3, If P = A, then JL^L = . and JLZliL^XlllL. if* - yi *2 - *i y* - yi If P ^ A, then by Theorem 1 1 .9 the slope of AT equals the slope of AS since AP and A B are segments (not necessarily distinct) of the same line aS. But the slope of £B is ^ ~ * jl and the slope of A? is ^~ & . x 2 — Xi x — Xj Therefore ya-¥i _ g-ffi X 2 - X, "" X - Xi " Multiplying both sides of this last equation by x — Xi, then dividing both sides by y 2 — y& then simplifying, we get * - *i _ y - yi *2 — %i 1/2 - yi " Therefore, in all cases, whether A = P or A=?=P t 'd P £ AB, then * - *i _ y - tfi ac 2 - xi y a - 1/1 Proof of 2: Suppose that P = (x, 1/} and that x-xi _ y- yi Xi - x, " y 2 - yi " Either * — Xi = or x — x± ^= 0. If x — xi = 0, then 1/ — j/ a = 0, (a, y) = (xi, yi), P = A, and P € AB, If x — x, ^£ 0, multiply both sides of *-*t _ y-yi b y 2 - yi *2 - xi Jfa — yi x - xi to get Uz-lfi y-Ui Xg — Xi X — Xl Since ~ 5EL is the slope of AB and *• *± is the sloi>e of AP, this #2 — *l X — Xi proves that if x — x x =j£ 0, then the slopes of AB and AT 5 arc the same and, as we shall prove, P Q AB. 484 Coordinates in a Plane Chapter 11 Suppose, contrary to what wc assert, that the slopes of A H u i id . \ / ' are equal, but that A } B, P are noncollinear as in Figure I. .1.-26. Figure 11-26 Let Qfe y') be the point in which the vertical line through P intersects AB. It follows from Theorem 11.9 that the slopes of A%) and AB are equal, so the slope of A$ is equal to the slope of AP, and y* -y\ _ y -yi X — Xt X — JTi * y' - ui = y - y^ ft = y> and the points A, B, P are collinear. Since our supposition that A, B, P are noncollinear leads to a contradiction, this proves that our suppo- sition is false. Therefore A, S, P are collinear and P lies on AB. This completes the proof in all cases, whether x — Xi =: or x — X\ y^ 0, that if * - *i _ y - yi *2 - *i "" y* - yi and P = fc y) t then F £ AB. This completes the proof of Theorem 11.14. The equation * - *i _ y yi yi- yi in the statement of Theorem 1 1 . 14 is often written in the form (i) y - yi = *±=3Ht - x,). *2 — 3.1 11.8 Other Equations of Lines 485 Show how Equation (1) is obtained from the one in the statement of Theorem 11.14. Since A = (acj, y\) and fi = {x^ y%l are distinct points on the oblique tine AB of Theorem 11.14, then -2i = m — *1 is the slope of AB. Substituting m for ^ ^ in Equation (1), we (2) rj - y t = m(x - xx) for AB. It should be noted that Equation (2) still holds if XS is a hori- zontal line since, in this case, m = and Equation (2) reduces to y = yi. Thus, if we know the slope of a line and the coordinates of any point on the line, we can write an equation of the line. When an equa- tion of a line is written in the form of Equation (2), it is often referred to as the point slope form. We have proved the following theorem, THEOREM 1L15 (The Point Slope Form) If Hs the line through A = (xu yi) with slope m, then '={(*> y) ' v - *j\ - »*(* - *i)} Example 1 If A a ( — 2, 3) and B = (4, ft), write an equation of Solution: Substituting the coordinates of A and B in the Two-Point Form of an equation, we obtain * - (-2) y-3 4 - (-2) 6-3 or x 4 L 2 _ y -3 6 ~ 3 as an equation of AB. IF we multiply both sides of the equation in Example 1 by 6, and add — 2y 4- 6 to both sides, we obtain x - 2y + 8 = t which is of the form Ax + By + C = 0, with A = 1, B = -2, and C = 8. 4-86 Coordinates "in a Plane Chaptarll This latter form is often referred to as the general form of a linear equa- tion, that is, of an equation whose graph is a line. Although we shall not do so here, it can be proved using the theorems of this chapter that if A, £, C are real numbers with A and B not both zero, then Ax + By + C = Ois an equation of a line. The converse statement is also true, that is, every line has an equation of the form Ar + By + C = in which A, B, C are real and A and B arc not both zero. What is the graph of Ax + By + C = if A a 0, B = 0, C=£0?ifA = 0\ B = 0\C = 0? Example 2 Write an equation of the line through (3, —4) with slope — ■§ and put the equation in general form. Solution: Substituting (3, — 4) for {%%, grj and — -| for m in the Point Slope Form of an equation, we obtain i, + 4 = -|<r - 3). Multiplying both sides of this last equation by 3 and adding 2* — 6 to both sides, we obtain 2i + 3i/4-6 = 0asan equation of the line in general linear form. Every nonvertfcal line intersects the y-soas (Why?) in a point whose coordinates are (0, h), where b is a real number. The number b is often called the y -intercept of the line. Similarly, every nonhorizontal line intersects the z-axis in a point whose coordinates are (a, 0), where a is a real number. The number a is called the x-mtercept of the line. It should be clear that die x- and {/-intercepts of a line can be obtained from an equation of the line as follows: 1. If the line is a nonhorizontal line, the ^-intercept is obtained by substituting for tj in an equation of the line and solving the resulting equation for x, 2. If the line is a tionverrical line, the (/-intercept is obtained by substituting for x in an equation of the line and solving the resulting equation for y. Example 3 Find the x- and t/-tntcrccpts of the line whose equation is 3s - 4y + 8 = 0. Solution: Substituting for y i u the equation 3x — Ay -f 8 = 0, wc obtain 3x + 8 = 0, or x = — 2$. Hence the ^-intercept is — 2|. Sub- stituting for x in the equation, we obtain — Ay + 8 = 0, or y = 2. Hence the {/-intercept is 2. If we substitute (0 T b) for {%i, ij\ equation of a line, we obtain in 11.8 Other Equations of Unas 487 the Point Slope Form of an or y — b = m(x — 0) tj = mx + b. It should be clear that when an equation of a nonvertical line is put in the form ;/ = mx + b, the slope and (/-intercept of the line can be read directly from the equation, This form, which is often called the slope y-intercepl form, is especially convenient when one wants to draw T tlie graph of a line whose equation is given. Exanipte 4 An equation of the line / is 3* + 2y — 8 = 0. L Put the equation in slope (/-intercept form. 2. Draw the graph of I on an xy-p\m\c. Solution: 1. To put the equation 3x + % — 8 = in slope (/-intercept form we add — 3x 4- 8 to both sides and then divide both sides by 2. The resulting equation is u = — Jar + 4. 2. The graph of t is shown in Figure 1 1-27. Note that the slope and (/-intercept of I can be read directly from the equation y — — Jx -(- 4, Thus, if we start at the point where y = 4 on the y-axis [that is, the point (0, 4}] and "trace out" a slope of — §, we arrive at a second point on /, These two points determine / and henee they determine the graph of I. I - V> ».4)1 5 2 C~ i - 3 V \ X -5 JV_ 5 Figure 11-27 As you might expect, two nonvertical lines are parallel to each other if and only if they have the same slope. If you worked Exercises 22 and 28 of Section 11.7, you should have discovered a relationship between tlie slopes of two lines that are oblique and perpendicular. We state these properties of parallel nonvertical lines and of perpendicular oblique lines as the last two theorems of this section. 488 Coordinates in a Plane Chapter 11 THEORESf 11.16 Two non vertical lines are parallel if and only if their slopes are equal. Proof; Let two non vertical lines r and 9 be given. We have two things to prove. L If r || $, then their slopes are equal. 2. If the slopes off and s arc equal, then r s. Proof of 1: Suppose that r and * are nonvertieal lines and that r [f $, If r = s T then r and s are the same line and hence they have the same dope. Suppose r^=s t as shown in Figure J J -28. Let l\xu yi) and Q(xt, 1/2) be two distinct points on r. Let the vertical lines through P and Q intersect s in F[xi, iji + k) and Q'{x 2> j/2 + k), respectively. Figure 11-2H Then PFQ'Q is a parallelogram (Why?) and PF = Q^. But PF = \h\ and QQ 1 = k\. Why? Therefore \h\ = \k\. Since h and h are either both positive or both negative, wc have h = L The slope of Pfp (and hence of r) is *2 — *i .The slope of FQ' (and hence of s) is (jfe + A) - (yi 4- h) _ y 2 -y 1 %2 — *! X2 - Xj ' since ft = fe Therefore the slopes of r and s are equal. Proo/ o/2: Suppose that r and s are nonvertieal lines and that their slopes are equal. If r = s ( then by definition r I .9. Suppose that r =£ s. Let m be the slope of r. Then s also has slope m. Now, either r and s arc 11.8 Other Equations of Lines 489 parallel Of they are not. Suppose they are not parallel. Then they have exactly one point P(a"i, r/i) in common. Tin is we have two distinct lines passing through the same point and having the same slope. This con- tradicts Theorem 11.10, Therefore our supposition that r and -s are not parallel is incorrect; hence it follows that r || $ and the proof is complete. THEOREM Il.il Two oblique lines are perpendicular if and only if the product of their slopes Is — 1. Proof. Let lj and Z 2 be two given oblique lines with slopes mi and wig, respectively. We have two statements to prove. 1. If /i _L h, then mi • mg = — 1 . 2, Umi*m 2 = — 1 , then h _ k- Before proceeding with the proof we wish to comment on the adjective "oblique" in the statement of the theorem. Would the state- ment that results if "oblique" is erased be a theorem? No, it would not. For if one of two lines is not oblique, then those two lines arc perpen- dicular if and only if one of them is horizontal and the other is vertical. Since a vertical line has no slope, there would be no product of slopes in this case. We now proceed to the proof of statements 1 and 2. Let p\ and p% be the lines through (0, 0) and parallel to h and fa, respectively, as shown in Figure 11-29. Then the slope of pj is mi and tie slope of p% is nio. Why? Since neither »i nor »2 is a vertical line, they both intersect the line I = (fc tj) : x = 1}, Figure 1 1-23 p\ 490 Coordinates in a Plane Chapter 11 Let A = (1, gft) and B = (1, 1/2) be the points of intersection of I with lines pi and p 2 , respectively, as shown in Figure ] 1-30, Figure 11-30 Tli en the slope of p\ is and the slope of p 2 is Wi -0 y 2 -0 m 2 = *— ~ = y 2 . Therefore A = (1, mi) and B = (I, ma). Now, h _L l 2 if and only if pi 1 pz- It follows from the Pythagorean Theorem and its converse (applied to AOAB in Figure 11-30) that Pl _ p 2 if and only if (OAf + (OB)* = (AB) 2 . From the distance formula, we get (QA)* = 1 4- wn 2 , (OB) 2 s 1 + mtf* and Thus (AB) a = (ms - mi)*. if and only if 1 + mi 2 ■+■ 1. 4- m 2 2 ss (mg — mi) 3 , if and only if 2 -j- w-i 8 + «»^ = m 2 2 — 2m im^ + »»i a , if and only if 2 = — 2f9t3!fft& if and only if mim? = — 1, Therefore £1 _L Z 2 if and only if m jm 2 = — 1, and the proof is complete. 11.8 Other Equations of Lines 491 Example 5 If h =((*,!,): 5*- 2y + 4=0}, h = ((jtv y) : 2a: + 5y - 15 = 0}, and fe ■ {(a& </) : 5* - 2# - 8 = 0}, show that l\ I ?3 and that ti ± h. Solution; Putting the equations of I it f 2 , /g in slope ^-intercept form, we get h - {{*> V) : </ = $* + 2}» I 2 = {(*,</): y = — $* + 3}> and k = {(J, y) : y = fx - 4}. Since ^ = slope of li = slope of h> it follows that 1$ \\ I3, Also, since the product of the slopes of l-\ and 1% equals *•(-« = -i. It follows that /i X fe, K »s also true that (3 J- k- Why? EXERCISES 11.8 1. Prove Theorem 11.13. In Exercises 2-6, use the Two-Point Form to write an equation of the line containing the given points, and put each equation in general form (that is, the form Ax + By + C = 0). 2. (1, 5) and (3, 4) 5, (<X 0) and ( - 1, 6) 3. (0. 3) and (-5, 0) 6. (-2, 2) and (2, -2) 4. (0 t -3) and (5, 0) In Exorcises 7-12, use the Point Slope Form to write an equation of the line which contains the given point and has the given slope, and put each equa- tion in general form. 7. (3, 5) and m = 1 8. (-2, l)andm= -1 9. (a0)andm = | 10. (5, 0) and m = -| 11. (-3, -7)audm = -1 12. (-5, -3)andm = | 492 Coordinates in a Plane Chapter 1 1 ■ In Exercises 13-16, determine which word, parallel or perpendicular, would make a true statement. 13. The lilies of Exercises 7 and 8 are |T] . 14 The lines of Exercises 8 and 1 1 lire [Tj and distinct. 15, The lines of Exercises 9 and 12 are [TJ . Are these two lines distinct? 16. The lines of Exercises 9 and 10 are [?]- 17* Write an equation (in general form} of the line which contains the point {3, 8) and is parallel to the line whose equation is 2x — 3y = 10. (Hint: What is the slope of the line whose equation is 2* — 3y = 10?) 18. Write an equation (in general form) of the line which contains the point ;3, Si and is perpendicular to the line whose equation is 2x — Zlj = 10. 19. Write an equation (in general form) of the line which contains the origin and is perpendicular to the line whose equation is y = x. ■ Iu Exercises 20-25, an equation of a line is given. In each exercise, (a) put the equation in slope (/-intercept form, (b) find the slope of the* line, and (c) find the x- and ^-intercepts of the line. 20. 2x - 3y - 12 = 21. 3x + 2y = 16 2& 4x - 6y = -8 23* y + x = 24. 4x - 2ij = 1 1 25. 5x + 4y + 13 m 26. Which pairs, if any, of the lines of Exercises 20-2-5 are parallel? Which pairs, if any are perpendicular? 27. Find, without graphing, the coordinates of the point of intersection of the lines P = {{*> a) : * + % = 6} and q={(x,y):5x + 4tf= -3}. (Hint: Note that p is not parallel to q (show this) and hence the lines intersect in exactly one point.) l>et (x u y±) be the point of intersection of lines p and q. Then, since (*j, j/j) is on p, x x + 3i/i = 6 or (1) Ui m - j*i + 2. Also since (x\, y^) is on q ¥ Sxj -f 4yi = —3 or m sn = -^ - 1 Apply the substitution property of equality to Equations ( 1} and (2) and find xi and tji. 1U8 Othar Equations of Lines 493 28. Show without graphing that lines p={(x > t,):2x-3y=i2} and q = fc y) - 4y + x m -5} are not parallel, and find the coordinates of their point of intersection, (See fjcercise 27.) Exercises 29-39 refer to the triangle whose vertices are A = (1, 1), B = {9, 3), and C = (7 t 9). In each case where an equation of a line is asked for, write the equation in slope (/•-intercept form. (It may help to draw the figure in an xy-plane and label the points as you need them.) 29. Find the coordinates of the midpoint D of AB. 30. Find the coordinates of the midpoint £ of BC. 31. Find the coordinates of the midpoint F of AC* 32. Write an equation of the line through A and & 33. Write an equation of the line through B and F. 34. Write an equation of the line through C and D, 35. Show that the lines A£. Bb\ CD intersect in the same point. 36. Write an equation of the line through E and F. 37. Write an equation of the line through A and B. 38. Show that F? || XS and hence that bW | AB. 39. Show that EF = \AB. 40. If a # and h s£ 0\ an equation of the form J- -f- = 1 is called the ah interrupt form of an equation of a line. Show that the line whose equa- X ft tion is r i — l contains the points (a, 0) and (0, b). 41. Put the equation 3* -f Ay = 12 in intercept form (see Exercise 40) and read the .v- and [/-intercepts directly from the equation. 42. Given that p = ((x t y) : x = $}, (a) Is(3 5 7)€p? (b)ls(3,l7)ep? (c) b(4,4)Cp? (d) Is (3, -9) € p? 43. Given that q = {(a-, y) : y = - i}: (a) Is(-4,5)€<j? (b) Is(-4, V5)€<?? (c) Ik(-4,-w)€tf (d)Is(4,-4)Cc/? 494 Coordinates in a Plane Chapter 1 1 44. challenge problem. Use theorems of this chapter to prove the fol- lowing statement: If A. B, C arc real numbers with A and H not both zeru T then Ax + By + C = is an equation of a line. 45. challenge problem. Use theorems of this chapter to prove the fol- lowing statement: Every line has an equation of the form Ax + By + C = in which A, B,C are real and A and B are not both zero. (Note that this statement is the converse of the statement in Exercise 44.) 11.9 PROOFS USING COORDINATES We have defined an xy-coordJnate system in a plane and have used coordinates as tools in much of our work in tills chapter, Given a plane, there are many xy- coordinate systems in that plane. In constructing a proof of a geometric theorem, it is wise to select a convenient ^-coor- dinate system that fits the problem and, at the same time, reduces the number of symbols needed in the proof. Such a selection yields no loss of generality, yet reduces the amount and difficulty of work involved. We illustrate with our next theorem which appeared as a corollary in Chapter 10. THEOREM 11.18 A segment which joins the midpoints of two sides of a triangle is parallel to the third side and has half the length of the third side. We shall give two proofs of Theorem 11.18. In the first proof, we select an arbitrary x ^-coordinate system in the plane of the triangle without any regard to the position of the vertices and sides of the given triangle. In our second proof, wc "pick" an ^-coordinate system in the plane of the Lriangle in such a way as to reduce the number of sym- bols needed in the proof, Proof I: Let AABC in any ary-plane be given. (See Figure 11-31.) Cfaun) figure 11-31 £(*S>«) 11,9 Proofs (Mill Coordinates 495 Suppose that A = (xi, yi) t B = (x 2j jfe), C = (x 3} y s ). Let D, E, F be the midpoints of BC y AC, KB, respectively. Then U.sinu; llic Distance Formula, we get (BE) 2 = (^_±ii _ *i ±j» V + / yg + ys _ g/i + ya \ 2 . = ft* - *i)) 2 + (Ksfc - ^)j*, But, by the same formula, {AB)* = {x 2 - x v f + (y, - m f. Therefore {DEf = i(AB)2 t ° r DE = #AB). Next* we prove that DE [| KB. Suppose that AB is a vertical segment Then, xj a 3C2 Rod — - — SB — i Therefore "~l 2 ' 2~~ /* F _ (*i + x 3 yi + y 3 \ and DE is a vertical segment. Hence DE || AB. If ZB is not u vertical segment, then its slope is ^ 2 ~ ^ 1 . The slope of TM is xs — Xi 2 2 _ y 2 - yi Xi + x s xi + x 3 x 2 - xi ' Therefore, since DE and AB have the same slope, DE [ KB. This com- pletes the proof so far as the segment DE is concerned. In a similar way, wc can_prove that EF = \BC and EF || BC and that DF = ^AC and DF II AC. 496 Coordinates in a Plane Chapter 11 Proof H: Let A A B C be given . 1 n the plane of this t dangle there is an ^/-coordinate system with the origin at A 9 with Afi as the x-axis, with the ^-coordinate of B positive, and with the ^-coordinate of C positive, (See Figure 11-32.) q2x&2>a) Figure 11-32 B{2xi,0) Let x lt x 2s y-2 he real numbers such that B = (2x [3 0), C — (2x2, 2f/2}* Let D t £, f* be the midpoints of EC, AC, AB, respectively. Then D = (xi + x s , i/ 2 ), E = (x 2s y 2 )» the slope of DE = 0, the slope of KB = 0, DE a |{X! + x 2 ) - x 2 | = jxj = x lt and AJ? = 12x41 =2*|. Therefore DE || KB and £»E a {A#. This completes the proof for UE and this is all that we need to prove, since DE might be any one of the three segments which joins the midpoints of two sides of the given triangle. To prove the statement of the theorem for the segment which joins the midpoints of two sides, we first label the triangle so that AC and CB are those two sides and then proceed as above. Thus each of the three parts of the proof uses a different coordinate system, but what we write in each case is the same. For example,. Figure 11-33 shows another pic- ture of the triangle shown in Figure 11-32. However, it shows a differ- ent xy-coordinate system and a different labeling of the vertices. Vertex C in Figure 11-32 becomes vertex A in Figure 1 1-33, A becomes B t and B becomes C, and we have made the x-axis look horizontal. You should note that the proof that DE = \AB and DE \\ AB would proceed exactly as before, but UR in Figure 11-33 is not the same seg- ment DE as in Figure 11-32. The same applies for the segment AB, 11.9 Proofs Using Coordinates **? C(2i 2 ,2>'2) Figure 11-33 It is clear Hi at Proof II is simpler than Proof I and is to be preferred. In general, if a proof using coordinates involves a polygon, it is usually easier to construct a proof if we select an ^-coordinate system in the plane of the polygon satisfying one or both of the following conditions: 1 . Let the origin be a vertex of the polygon and let the positive part of the x-axis contain one of the sides of the polygon, 2. If the polygon contains a right angle as one of its angles, let the origin be the vertex of the right angle and let the positive parts of the x- and (/-axes contain the sides of the right angle. THEOREM 11,19 The medians of a triangle are concurrent in a point (eentroid) which is two-thirds of the distance from each ver- tex to the midpoint of the opposite side. Proof: Let A ABC he given. Select an ^-coordinate system in the plane of this triangle with the origin at A, with AB as the x-axis, with the abscissa of B positive, and with the ordinate of C positive. (See Figure 11-34.) Let a, h, c be numbers such that B = [Ga, 0), C = {&b f 6c), Let D s E, F be the midpoints of BC, CA, X73, respectively. i * cm*;) E(3b>3e/ J \Dr3o+36,3e> / Nt'" A /uo.0} i-\3c.U) 0(60,0) Figur# U-34 498 Coordinates in a Plane Chapter 11 Then AD, BE, CFare the medians of A ABC. We must prove that AD, BE, CF are concurrent at some point P and that AF = JAD, BP m |BE, and CP = |CF. The midpoint of AB is F(3a, 0). We can express CF parametrically as follows: QP= {fcy) : x = 66 + {3a - 66)*, y = Gc + (0- 6c)*, < * < 1}. The point P on CF such that CP = |CF can be obtained by setting k = | in the parametric equations for CF. Thus x = bh + (3a - 66} • J - ($6 + 2a - 46 = 2b + 2a and Therefore y = 6c + (0 - 6er) ■ § = 6c-4c = 2c. P= (2a + 26, 2c). Similarly, the midpoint of J5C is D = (3a + 36, 3e) and AD = {{s, • i = 4. (3c + 36 - 0)fc, i/ = + (3c- 0)*,0< fc < 1}. The point F on A D such that AF = %AD is obtained by setting k = -J . Tims s= (3a + 3/?)* | = 2« + 26 and f -(*)*} = 2c. Therefore F = (2a + 2b, 2c). The midpoint of CA h E = (36. 3c) and BE = (fey) : x s 6a 4- (3fr - fia)*. y = + (3c - 0)*,0 < k < 1}, The point F' on BE such that BF' = §BE is obtained by setting k = § . 11.9 Proofs Ustnf Coordinates 499 Thus and Therefore x = 6a + (36 - 6fi) ■ | = 6a + 2fo - 4c = 2a + lb = 2c. P" = (2a + 2fc, 2c). Wc have shown that P = r = P" = (2a + 26, 2c), that CP = fCF, AP = \AD t and that BP = %BE. Therefore the medi- ans of a triangle are concurrent at a point which is two- thirds of the distance from each vertex to the midpoint of the opposite side, and trie proof is complete, In the second sentence of this proof we could have taken a t h t c as real numbers such that B = (a, 0) and C = {b t c). The resulting ex- pressions for the coordinates of D, E T l\ and P would have involved many fractions. We avoided these fractions by taking a, 6, G so that B = (0a, 0) and C = (65, 6c). You may feel that Theorem 11,19 would be easier to prove without using coordinates. It is possible to construct such a proof using the the- orems, postulates, and definitions that we have established before this chapter. You might be interested in trying to do so. As Indicated in the statement of the theorem the point of intersec- tion of the medians of a triangle is its ceutroid. In informal geometry we think of it as the balance point; it is the point where a cardboard triangular region of uniform thickness balances. In calculus the idea of moments of mass (extending the idea of weight times distance in teeter-totter exercises) is introduced and extended to develop a theory of centroids for plane figures. The centroid of a triangle is an example of a centroid as the concept is developed formally in calculus. Our last example of this section is a theorem that you will find help- ful in working some of the exercises at the end of the section. THEOREM 11.20 Let quadrilateral ABCD ivith A = (0, 0), B = (a, 0), D = (h> c) lie given. ABCD is a parallelogram if and only if C = (a + b> c). MM Coordinates in a Plane Chapter 11 Figure 11-35 shows one possible orientation of Lhe given quadri- lateral ABCD in an xt/-plaiic. However, our proof depends only on A being at the origin and 8 being on the it-axis as given in the theorem. We must prove two tilings. h JfC = [a + b, c), then ABCD is a parallelogram. 2. If ABCD is a parallelogram, then C = (a 4- b, c). D(b t e) a**y) ^m 5:q.,u; Figure 11-35 Proof of h If C = (a + b, c), the slope of CD is c — c _ 0. = o a + b — b a Also, the slope of AB = ^ = 0, Therefore CD \\ AB. Since KB and UD ai"c horizontal segments (Why?}, we have AB = \a\ and CD = \a + b - b\ = a\. Therefore AB — CD and ABCD is a parallelogram. Proof of 2: If ABCD is a parallelogram, we must prove that C = (a + b r c). Since ABCD is a parallelogram, KB || t25. The slope of A J? is 0j therefore the slope of CD is 0. Let C = (x, y); then the slope of CD Ls = 0. a; — ITierefore y-c = 0andy= c. We have AB = CD (Why?) and AB = |d| and CD = \x - b\. Therefore \a\ = k - R 11.9 Proofs Using Coord f nates 501 Now, if a > 0, then x > b % and if a < 0, then * < b. (If a > and x < fc, or o < and x > fc t then C and B would be on opposite sides of AD and ABCD would not be a parallelogram.) It follows that a = x — b or that ac = a ■+■ fc. Therefore C — (a + h, c) and the proof is complete. EXERCISES 11.9 Unless stated otherwise, use coordinates to prove the theorems in this set of exercises* Many of these theorems have appeared as theorems or exercises earlier in the texL We include them here since they can be proved easily using coordinates. 1. Prove: THEOREM 11.21 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram, (Hint: Let A = (0, 0). B = (o, 0}, C = (x, y), and D = {h, c) be die vertices of the quadrilateral and suppose die diagonals of the quadrilateral bisect each other. Show that x = a + b, y = a, and then apply Theorem 11,20.) 2. Prove: THEOREM 11.22 The diagonals of a parallelogram bisect each other. (This is the converse of Theorem 1 1.21. By Theorem 11 -20, the vertices of a parallelogram may be taken as A = (0, 0). B = (a t 0), C = (a + h, c), and D = (h, c)+ Show that the midpoint of AU Is the same point as the midpoint ,,] HI I) X Prove: THEOREM 77.23 If the vertices of a parallelogram are A = (0, 0), B = (a, 0), C =; (a + b t c), D = {b, c), then the parallelogram is a rectangle if and only if b = 0. (You must prove ( 1) if ABCD is a rectangle, then h = and (2) if b = 0, then ABCD is a rectangle.) 4. Prove: THEOREM 11.24 The diagonals of a rectangle are congruent, (Let ABCD be the given rectangle. V$e Theorem 11,23 and prove AC m BD.) 502 Coordinates in a Plane Chapter 1 1 5, Justify the steps in the proof of the following theorem. THEOREM 11,23 If the diagonals of a parallelogram arc congruent, then the parallelogram is a rectangle. Proof: Let A = (0, 0), B = (a, 0). C = {a + b, c), D = (b t c) be the vertices of the given parallelogram. What theorem justifies our writing C = (a + fc, c)? We have Therefore and Therefore AC = BD. (AC)* = (BD) 2 , (AC) 2 = (a + &)» + ^ 2 > («Z))8 = {& - o^ + c 2 . ( fl + b) 2 + c 8 = (J> - «)2 + c a . Simplifying, a 2 + 2<ii> + L* 4- c 2 = b 2 - 2afe -f- a 2 + c\ 2ab= -2ab t and 4«/> = 0, Staee 4a 7^ 0* we can divide Ixith sides of the last equation by la, ob- taining h = 0. Therefore ABCD is a rectangle. 6. Justify the steps in the proof of the following theorem . THEOREM 11.26 A rectangle is a square if and only if its diagonals are perpendicular. (See Figure 11-36.) By Theorem 11.23 we may write A = {0, 0), B = (a, 0), C = (a, c), D = (0, c) for the vertices of the given rectangle. There are two things to prove. 1. If AT X SB, then ABCD is a square. 2. If ABCD is a square, tlicm A~C _ WD, Proof of 1: We arc given AT 1 UD. The slope of AC h r mid the slope of #D is -£—, Therefore — a — a 11.9 Proofs Using Coordinates 503 Therefore c 2 - = _l ( fiS — fl a t an d f c f _ u , t -a* But |c| = BC and |« = Afl. Therefore BC = AB and A£GD is a square. Proof of 2: We ate given ABCD is a square. Therefore Afl = BC; \a\ = \c\ f and a- = c 3 , I lit r»j be the slope of AC and m-> be the slope of BD. Then Wi = — , 'Us = — -. and mirn^ = — -Q-, But — = l s .so mim? a — 1, Therefore W -I BB. 7. Justify Ihc steps in the proof of the following theorem. THFOBKV 11.27 If the vertices of a parallelogram are A = {0, 0), B = (a, 0), C = (a + &, c), and D = {&, c), then the parallelogram is a rhombus if and only if a s = &+ c*. There are two things to prove, 1. If a 1 = b s + c* then ABCD is a rhombus. 2. If ABCD is a rhombus, then flS = 6 2 - A Proof of 1: We are given that a 2 = A* + e* so a V&* + C*i By tlie Distance Formula, ADm y/& + A Therefore AD = |of. Also, AB = a\, Therefore AB = AD and ABCD is a rhombus. Proof of 2; ABCD is a rhombus, so AD = AB and Therefore and the proof is complete- 504 Coordinates in a Plans Chapter 11 8, Prove: THEOREM 11.2H If the diagonals of a parallelogram arc perpen- dicular, then the parallelogram is a rhombus. (Hint: Let A = (0, 0), » = {a, 0). C = (a + b % c), and D = (6, c) be th e ve rtices of the given parallelogram. l,et mi and m% be the slopes of AC and BD, respectively. Using Theorem 11.27 of Exercise 7 and mi • m-t = — I , show that a 2 = b 2 -+- cK) 9. Recall that a trapezoid is a quadrilateral with at least one pair of parallel sides which are called the bases of the trapezoid. The other two sides are called the legs of the trapezoid. The segment joining the midpoints of the legs is called the median of the trapezoid. Prove the following theorem. THEOREM 11.29 The median of a trapezoid is parallel to each of the bases and its length is one-half the sum of the lengths of the two bases. [Hint: Let A = (0, 0). B = {2a, 0), C = (2a\_2c), and_D a (2b f 2c) be the vertices of the given trapezoid. Then AB and CD arc the parallel bases. I*et E and F be the midpoints of AD and BC r respectively. Show that EF I AB t ET || CD, and that EF = fy\B + CD).) 10. A trapezoid is isosceles if its legs arc congruent (See Exercise 9.) Prove the following theorem, THEOREM 11.30 A trapezoid is isosceles if its diagonals are congruent 1L Prove: THEOREM 11.31 If a line bisects one side of a triangle and is parallel to a second side, then the line bisects the third side of the triangle. 12. Prove: THEOREM 11.32 The midpoint of the hypotenuse of a right triangle is equidistant from the vertices of the triangle. (Hint Let A = (0» 0), B m (2a, 0), and C = (0, 2fc), where a, b are positive numbers, be the vertices of the given rigjht triangle.) 13. Complete the proof of the following theorem. THEOREM 11.33 The lines which contain the altitudes of a triangle are concurrent. (Their common point is called the ortbocculer of the triangle.) (Sec Figure 11-37), Froof: Let A = (a, 0), B = (b, 0), C = (0, c)> and suppose that a < b t < c as shown in Figure 1 1-37. I jCI A', B\ C be the feel of the per- 11,9 Proofs Using Coordinates 505 i kj sm t / -\ \ * \.mo) i. * * t t / Ugam IU1T pcndfculars from A, B, C to BC, CA, AB, respectively. You arc to prove that AA', BB', CC' arc concurrent at some point F{xi, y{), (Note that (HO, 0) is the origin.) The slope of BC is b c-0 0-h •v — r ■* r — — y. Since .'LA' 1 EC, the slope of A A' is — (Why?), Using the point slope form of an equation, we may express A A' as +_+ * AA' = {(x,t 7 }:i/=:^-a;.}. * — > 4 — > since (a, 0) is a point on AA'. We may express CC as CC={(x 1 tj) :* = <)}. Now write an equation for BB\ Let P(*i, 1/1) be the point of inter- section of lines AA', CO, and let F[x2, </?) b* f he point of intersection of lines BB', CC, Solve the equations for AA', CC "simultaneously" to find P[xi, t/i) an<l solve the equations for BB', CC simultaneously to find P(x 2 , ?/2)- Show that P = F and hence the lines AA\ BB'. CC are concurrent at P. 14. Prove Theorem 11.18 by use of definitions, postulates,, and theorems studied before this chapter: that is, without the use of coordinates. (Perhaps you did this in an exercise of Chapter 10.) 15, Use Theorem 11,38 and theorems, definitions, and postulates studied before this chapter to prove Theorem 11.29 without the use of coordinates. 16* challenge Pitoni.KM. Prove Theorem 11.19 without die use of coordinates. 506 Coordinates in a Plane Chapter 11 CHAPTER SUMMARY In this chapter we used the idea of a coordinate system on a line to de- fine an .Tj/-coordinate system in a plane. We showed that there is a one-to- one correspondence between the set of all points in a plane and the set of all ordered pairs of real iwmlwrs. The key theorems in this chapter are: THEOREM 11.4 If F| = (x u yi) and P? = (** y 2 ) are any two points in an .vj/-plane, then PiPs = V(*i - «a) B + (m - W< THEOREM 11.5 If P = (x h ijt) and Q = (x 2 , ij 2 ) are any two distinct points in an jt]/-planc, then the midpoint X! of PQ is the point \ 2 ' 2 r TlIEOR EM 11.7 UA(xutfi) and Bix-t, ys) are any two distinct points, then AB = {x, y) : x = x t + k(xt - *i), y = y, + %o - y^fcisreal}. If jt is a real number and if F = (x, y) where ar = Xi + fe(*2 — Xi), u = yj - %a - yi)» &<»» AP = *(ArJ) and P^AJ? if it > 0; AP = _*(AJ9) and P£oppAR if Jt < 0- The formula in Theorem 1 1.4 is called the DISTANCE FORMULA. The formula in Theorem 11*5 is called the MIDPOINT FORMULA. The equations x z= x% + k(x$ — Xi) and y = tji ■+- fe(j/ 2 — #i) in The- orem 11.7 are called FARAMETRIC EQUATIONS for the line A~S, and k is called the PARAMETER. We proved that the converse of Theorem 11.7 also holds; that is, if tt, h, a, d are real numbers, if b and d are not both zero, and if S = {{x* y) '. x = a + bk, y = c + <&-, fc is real), then S is a line. We defined the SLOPE of a non vertical line to be the slope of any one of its segments. We defined the slope of a non vertical segment with end- points Pi(*i, t/i', 'Wt**! Vs) *° l** 5 ■ We showed that two non vertical lines are parallel if and only if they have equal slopes and that two oblique lines are perpendicuilar if and only if the product of their slopes is — 1. Slope is not defined for a vertical line, but the slope of a hori7.ontal line is zero. We proved that if P|(*j, y{f and Psfra, y?) are any two distinct points on a nonvertical line, then * - *i _ V - tfi *a - *i J/2 - y\ Rsvlew Exercises 507 is an equation of the line. We called this form of equation the TWO-POINT FORM for an equation of a line. We proved that a line with slope m and passing through H x i> tf i) has an equation of the form y - (/t = m(x - *i) and called this form the POINT SLOPE FOHM for an equation of a line. We proved that a line with slope m and {/-intercept h has an equation of the form y s= mx + b and called this form the SLOPE (/-INTERCEPT FORM for an equation of a line. We showed that an equation of a vortical line through ¥(%{, r/i) is X — K\ and that an equation of a horizontal line through P{x\, iji) is y = y\. Wc called the form Ax 4- By + C = the CpENERAL FORM for an equa- tion of a line. Finally, we showed how coordinates could be used to construct proofs for some geometric theorems and observed t hat, in some cases,, proofs using cord (nates are easier than those using previously established definitions, postulates, and theorems, REVIEW EXERCISES Graph each of the sets indicated in Exercises 1-10. 1. {{*, y):x = 5, < y < 5} 2. [(*f):r- -3, -2<C*<7} 3. {(x, y) : x = 1 + 2Jfe, y = 2 + 3&, k is real} 4. ((* y) : x = 1 + 2*, y = 2 + 3*. k £ 0} 5. {(x, y) : x = -2 + fc, i/ = 1 - 2k k < 0) 6. {(a*, t/) : x = *, y = 3 - 2k, 2 < k < «} 7. {{*,«,) : l<x<4or2<y<5} 8. {(x, y) i 1 < s < '1 and 2 < y < 5} 9. {(x,y):y>2} 10. {{x, y) : y = |x + 5} In Exercises 11-13, the endpoints A and R of a sequent AB are given. Find (a) the slope of AM (in lowest terms), (b) the midpoint of SB, and (c) the dis- tance AB. 11. A = (2, 5} and ii = (-2,3) 12. A - ;-L -3) and B = (2, -9) 13. A = (2, -5) and B = (7, 7) 14. A = (-2, 1) and B = (2, 3) 15. A = (-3, l)andB= (7,1) 508 Coordinates in a Plane Chapter 11 16. Which segments in Exercises 11-15 are (a) parallel? (b) perpendicular? (c) congruent? 17. Write, in general form, an equation of the line which contains the points P=(-h ^3) and = (2, -9). IS* Write, in general form, an equation of the line which has a slope of ^ and contains the point H = (—2, 1). 19. Prove that the line of Exercise 17 is perpendicular to the line of Exercise 18. 20. Write., in slope (/-intercept form, an equation of the line with slope J arid ty-intercept 6, ■ In Exercises 21-25, an equation of a line is given. In each exercise, (a) put the equation in slope {/-intercept form, (b) write the slope oi' the line, and :::) write the x- and {/-intercepts of the line. 21. 3x + 2y m 12 22. 2x - y = 7 23. lox ~2lij = 7 24. x + y = 25. x — y = 86, Stow that A SKM is a right isosceles triangle if S = {3, 4), X = { - 1, 3), andM = (-2, 1). 27. Given A = (1, 0), B = (4, 3), express AS using set-builder notation and parametric equation*. 28. Given A and B as in Kxerciso 27, express AB using set-builder notation and parametric equations. 29. Given A and B as in Exercise 27, find the trisection points of AB r 30. Write an equation of the vertical line through (2, 5). 31. Write an equation of the horizontal line through % 5). 32. Write an equation of the line through (3, —7) and parallel to the line with equation y = 3x -f 5, 33. Write an equation of the line through (3, —7) and perpendicular to the line with equation tj = 3x + 5. 34. Given the line with equation 5x - % = 00. write the equation of this line in slope {/-intercept form. 35. Given the line of Exercise 34, write the equation of this line in intercept form. 36. Given p = {(x, y) : x = 3} and q = {(x, y) : 2x = 6}, explain why Revww Exercises 505 Exercises 37-47 refer to the rectangle PQRS whose vertices are P = (-1, -4), Q = {5, -4), R = (5, .3), S = (-1, 3;. 37, Fmd die midpoint oi PR. 38, Find the midpoint of SQ. 39, Show that PQ = $R r 40, Show that PR = SQ. 41, Write parametric equations for PQ, 42, Write parametric equations for PR. 43, Find A on Pit such that PA = 4PR. 44, Find B on PR such that PB = fPR. 45, Find C on opp M such that PC = J/»«. 46, Write parametric equations for the line through 5 and perpendicular to PR. 47, Write parametric equations for the line through R and parallel to SQ. 48, Let die trapezoid ABCD have vortices A = (0, 0), B = (2a, 0), C = (2(1, 2c\ and D = (2/?, 2c), as shown in the figure. Let a, b, c, d be positive numbers such that b < d < a. Let E and F be the midpoints of AD and BC, respectively. Let EF intersect AC at P and SD at {). Show that P is the midpoint of AEL that Q is the midpoint of ED, and that P£ = {{AB - CD). P(M),2e) C(2d,2e) £(2b,0) 49. Prove, using coordinates, that the set of all points in a plane equidistant from two given points in the plane is the perpendicular bisector of the segment joining the given points. (Hint: Let the two given points in an xy-plane be A = (— a r 0) and B = (a, 0). Then the {/-axis is the per- pendicular bisector of AB in the given .try-plane.) There arc two things to prove. (a) If P(x, y) is on the y-axis, then AP = PH. (b) If AP = PR and P is in the .ty-plane, then F is on the y-axis. 50, challenge piiohlum. Prove that the urea S of a triangle whose ver- tices are A = (x\ t tj\), B = (X2, y2), and C =a [xz, y$) is given by the formula s = i^i^a + *»03 4- * 3 yi ~ *lft - x *y* - *fc§fll* Bradley Smith/Photo Researcher!, Coordinates in Space 12.1 A COORDINATE SYSTEM IN SPACE In Chapter 3 we introduced the fundamental idea of a coordinate system on a line, or a line coordinate system, or a one-dimensional co- ordinate system, as it is sometimes called. Tn Chapter H, we defined a coordinate system in a plane and called it an *j/-coordinate system. An any-coordinate system is a one-to-one correspondence between all the points in a plane and all the ordered pairs of real numbers. Each point has two coordinates. An ^-coordinate system Is a two-dimensional co- ordinate system. In this chapter we introduce the idea of a coordinate system in space and called it an xi^-coordinate system. In this system each point of space is matched with an ordered triple of numbers. It is •a three-dimensional coordinate system. Let a unit segment and die distance function based on it be given. All distances will be relative to this unit segment unless Otherwise indicated. Let OX and OY be any two perpendicular lines and let OZ be the unique line that is perpendicular to each at their point of Intersection, Let J, J, K be points on OX, OY, OZ % respectively, such that 01 = OJ = OK = 1. 512 Coordinates in Space Chapter 12 On OX there is a unique line coordinate system with O as Origin and / as unit point. On Or there is a unique line coordinate system with O as origin and / as unit point. On O'Z there is a unique line co- ordinate system with O as origin and K as unit point. We call these coordinate systems the x-eoordinate system on OX, the {/-coordinate system on OY t and the ^-coordinate system on Sz. We refer to OA\ OY, OZ as the x-axis, the (/-axis, and the *-axis, respectively We refer to them collectively as the coordinate axes. The plane containing the x- and y~axe$ is called the xy plane, The plane containing the x- and s-axes is called the j/z-plane. The plane con- taining the y- and s-axes is called the r/z-plane. We refer to these three planes collectively as the coordinate planes. From die theorems regarding parallelism and perpendicularity in Chapter 8 it should be clear that all lines parallel to the .s-axis arc per- pendicular to the K^-plane, that all lines parallel to the (/-axis are per- pendicular to the xz-plane, and that all lines parallel to the X-axis are perpendicular to the 1/2-plane, It should also be clear tit at all planes parallel to the xy-plane are perpendicular to the z-axis, that all planes parallel to the xz-plane are perpendicular to the [/-axis, and that all planes parallel to the i/s-plane are perpendicular to the x-axis. Figure 12-1 suggests the x-, i/-, and ^rcoordinate systems, Hie parts of tlie axes with negative coordinates are shown by dashed lines. They are not "hidden" from view by the coordinate axes in the figure. However, they are hidden from view by die coordinate planes, "llius the negative part of the x-axis is behind the i/s-pkne; the negative part of the c/-axis is hidden by the ace-plane; the negative part of the c-axis is hidden by tlie X(/-planc, In drawing pictures you should use your own judgment about whether a dashed segment is better than a solid one. Figure IS- 1 12.1 A Coordinate System in Space 513 In Figure 12-1, the positive parts of the x-, y- 3 and £-axes are ar- ranged as the thumb, forefinger, and middle finger, respectively, of a right hand when it is held as suggested in Figure 12-2* An .ti/5-coordi- natc system with axes oriented in this manner is called a right-handed coordinate system. If the unit points on the axes are selected so that the positive parts of the x-, y-, z-ax&s con form to the orientation of the thumb, forefinger, and middle finger of the left hand when the thumb and forefinger are extended and the middle finger is folded, the xyz- coordinate system is called a left-handed coordinate system. The fig- ures in this book arc for a right-handed system. Figure 13-2 There is an xy-coordinate system in the xy-plane determined by O, /, / as in Chapter 1 1 . This system is a one-to-one correspondence between the set of all points in the xy-plane and the set of all ordered pairs of real numbers. Similarly, there is an xs-coordmate system in the xs-plane and a ys-coordinate system in the y^-plane. T^et F be any point in space. Figure 12-3 shows P as not lying on any of the coordinate pianos. However, the following discussion which leads to the definition of an xf/z-eoordinate system applies to any point. For special positions of P some of the labeled points thai are distinct in Figiire 12-3 may not be distinct. For example, P and F IU may be the same point. Figure 12-3 514 Coordinates in Space Chapter 12 Let P w P«, V vt be the projections of P on the xt/-plane, the xs-plane, and the i/s-plane, respectively. I-et a XUt a gs , a^ be the planes through P and parallel to the xy-plane, the xs-plaiic, and the i/z-plane, respee- lively. Then a rv contains P t P rs , and P us ; fife* contains P, F m and F m % and %, contains P, P Mt and /', ••..,. Let @ n py, () t be the points in which a lw , a X!t a^ intersect the x-axis, the y-axis, and the s-axis, respectively. We are now ready to set up an .rr/s-eoordinate system. The ^coordinate of P is the ^.--coordinate of Qy, the y-coordinate of P is the ^-coordinate of Q^ the s-coordinate of P is the ^-coordinate of Q z . Wc write P = (a, 6, c) or P(a, Z? } c) to indicate that the *-, y-> s-coordi nates of P are a, b, c, respectively. If (a, h } c) is any ordered triple of real numbers, then there is one and only one point P such that P=P( a ,b>c), It is the intersection of three planes, one parallel to tbe xs-plane and cutting the *-axis at the point whose x-coordinate is a, etc. 'The cor- respondence between the set of all ordered triples of real numbers and the set of all points is a one-to-one correspondence. For if {«, b, c) and (rf, e t f) are different triple*; of numbers, then one or more of the following inequalities must hold: a^d, 6#e, c=£f. Suppose, for example, that a =£ tl Then P(«, b t c) and R(d> e, f) He in distinct planes parallel to the i/z-plane and therefore P^R. Definition 12.1 Given an t-axis, a (/-axis, and a zf-axis, the one-to-one correspondence between all the points in space and all die ordered triples of real numbers in which each point P corresponds to the ordered triple (a, h, c) where a, b, c are die Xh, y-, ^coordinates, respectively* of P is die xyz- coordinate system, Since the correspondence between points and triples is one-to-one, a system of names for the triples is a suitable system of names for the points. Thus (5, 6 r -3) is an ordered triple of real numbers. It is also a point. It is the point whose x-, y- f and s-coordinates are 5, 6, and —3, respectively. 12.1 A Coordinate System in Space 'j l :■ Example 1 If A = {(*, y, z) : x = 2. tj - 3, * = 4}, then A m {(2, 3, 4)}, that is, A is the set whose only element is the point (2, 3, 4). Example 2 If B={{x t t,,z):x = -2.y = 3}, then J3 is the set of all points (a; y, 5) such that the x-coordinate is —2 and the y-coordinate is 3, that is, £ is the line through ( — 2, 3, 0) and parallel to the .-7 -axis. Example 3 If C = {{*, y, z):y = 7}, then C is the set of all points whose ^-coordinate is 7, that is, C is the set of all points that are 7 units to the right of the acs-plancj hence C is the plane that is parallel to the xz-planc and 7 units to the right of it. Example I D = {(*, y, z) : x = 2 and y = 3}. The use of "and" !n this set-builder notation is the same as the use of a comma between statements of conditions in a set-builder notation. In other words, D = i(x i y i z}:x = 2,y = 3} is the set of all points (2* 3, z\ that is, the line parallel to the s-axis and passing through (2, 3, 0). Example o E = {(x, y, z) : x = 2 or y s 3}. Clearly* E is not the set D of Example 4. If you do not understand dearly the distinction between the use of "and" and "or" in a set- builder notation, you shoxild review Chapter 1. E is the set of all points with ^-coordinate 2 or ^-coordinate 3, hence the set of all points lying either in the plane x m 2 or in the plane y = 3. Tn other words, £ is the union of two planes, each parallel to the s-axis, one of them parallel to the yx-planc and 2 units in front of it, the other parallel to the xz-plane and 3 units to the right of it, assuming that the axes are as in Figure 12-1, 516 Coordinates in Spac* Chapter 12 Example 6 F = {(x, y,z):x=l + Zk t y = 2^ 3k t k is real}. Now {(*, y) : x = 1 + 2k, y = 2 + 3k, h is real} is a line I in the xt/-planc with slope | thai pisses through (1> 2). The set-builder notation for F places no restriction on the ^-coordinate. Therefore, if {x, y) is any point on I and z is any real number, then (x, y, z) is a point of ft Conversely, if (*, y, z) is any point on F, then (x, y) is a point of /. Think of /' as the set {(a; j/,2);a;=l + 2*,j/ = 2 + 3k, z = 0, k is real}. Then V is the same set as I If we think of a point of I, we think of it as a point (x, y) in an xy-plane. If wc think of the same point as a point of /'» we think of it as a point (x t y, 0) in an xip-spacc. Although the names of the points are different, the sets I and I' are the same. The graph of the set V is shown in Figure 12-4, Figure 12-4 Now V is a part of ft Indeed, F is the set I' and all points directly above it and all points directly below it. It may be helpful to think of it as the union of all linos parallel to the *-axis that pass through a point of V. But any way you look al F t you really have not seen it unless you recognize it as a plane, the piano through /' and parallel to the *-axis. Example 7 G = {(x t y,z):z<3}> G is the set of all points (*, y, z) whose z-coordinate is less than 3. Now z = 3 is an equation of the horizontal plane, say a, that is 3 units above the xc/-pkne. Then C is the halfspace that is the underside of a. 12.1 A Coordinate System In Spact 517 8 H=[(x t y 1 z):l< i z< 3}. H is a slab or zone of space bounded by two horizontal planes. H is the union of the horizontal planes with equations 5=1 and z = 3 and all of the space that lies between them. Example 9 I 8 {(*, tj,z):x* + y* = 25 t z = 0). 1 is a cirde in the xy-plane. Its center is the point (0, 0, 0). Its radius is 5. Example 10 / is a solid right circular cylinder. Its axis is a part of the s-axis. Its radius is 5. Its lower base lies in the xi/-plane. Its height is 6. Example 11 K = {(*, y, z) : x = 2 and x = 3}. There is no number x such that x = 2 and x = 3. Therefore there is no point (x, \j, z) such that x = 2 and x = 3. Therefore K = 0, the null set. Example 12 L = {(*, y, z) : x* - 3* + 2 = 0}. Now x> - 3* + 2 = is true if and only if (x - l)(x - 2) = 0; hence if and only if x = 1 or x = 2. Therefore L is the union of two planes, each parallel to the t/z-plane, one of them 1 unit in front of it, Ihc other 2 units in front of it. EXERCISES 12.1 In Exercises 1-35, a set S of points is given in set-builder notation In each case, descrilw the set S in words, assuming that the axes appear as in Figure 12- 1. {Hint: For Exercises 20, 21. and 22, compare with Example 12.) L $={{x t y.z):x = 0) %S={(x,y.z):y = Q} 3. S = {(x, y. z) i z = 0} 4. S= ((x,y,2):x = 0,y = 0) 518 Coordinates in Space Chapter 12 5. S = {<x s y, 4 19 m %* = 0) 6. S= {(x,y, £ )ry = 0»z = O) 7. S = {(x, y, s) : x = Q, y = Q, s = 0} 8. S={{x,y,- S >;x=l,y = 2,z=2} 9. S = {(*, y, z) : x = ly = 2, ] < * < 5} 10. S={(*,y,z) : a: = I, 1 <y^3,* = -I) It S = {(x, y, 2) : * £ 1, y = 3, z = 1} 12. S = {(*, y, s) : y < 3} 13. S = {{x, y, z) : x = 3, tj < 3} 14 t S = {(x, y, 2) : 1 <, x < 2, 3 < y <, 5, -1 < z < 3} 15. S = {{*, £/, z) ; x £ 0> y > 0, * > 0} 16. S = {{x, y, z):x> 370} 17. S = {(x, y,z):x = ly = 2) 18. S ss {(x, y, jj) ; x = 1 or J/ = 2} 19. S= {{x,y r z) : x = 5orx = 7} 20. S = ((x, y, 2) : s* + 3s + 2 = Oj- 21. S = {(x, y, z) i & + 2* + 1 = 0} 22. S = {(x, y, s) : x 2 = 16} 23. S={(x,y,s}:x 2 + y 2 = 25) 24. S = {(x, y, z) : x 2 + y* = 5, * - 5} 25. S= {(x,y, s ) :i/ = * + 3,5 = 4} 26. S = {(x, y, s) : tf s x} 27. S = ((x, y, *) : z = y} 2S. S = [<x, y, z) : 3x + Ay = 12, x £ 0, y > 0, 1 < 2 < 2} 29. S={(^, / ^):^ll == J^L? tS = 0} 30. S = {(x, y, z) ■ 2±1 = £z-L. y = } 3L S = {(x r j/,*):.t = 1 + 2k, y = 2 - ft, z = 0, 1 < Jfc < 2} 32. S = {(x, y, z) : x = 1 + 2*, y = 2 - Jfc, 1 < * < 2) 33. S = {[x, y, g) : x= + y* ^ 2 5, s = 3) 34. S = (fx, y, z) : y 2 + z* < £5. x = 3) 35. S = ({x,y,s):x#0.y.^O\s ? feO} ■ In Exercises 36-50, use set -I milder notation to express the set 5. 36. 5 is the xy-planc, 37. S is the xs-plane. 35. S is the j/*-p1ane. 39. S is the x-axis. 40. S is the y-axis. 12.2 A Distance Formula 519 41. S 42. S 43. S 44. S 45. S 46. S 47. S 48. S 49. S 50. S is the plane through (2, 6 f 7) and parallel to the yz-plane. is the plane through (2, 6, 7) and parallel to the xs-plane. is the plane through (2, 6, 7) and parallel to the xy-pfane. is the line through (2, 6, 7) and perpendicular to the t/2-plane. is the line through (2, 6, 7) and perpendicular to the xs-plane. is dio line through (2, 6, 7) and perpendicular to the xy-plane. is the line in the xr/-plane which contains (2. 3, 0) and (3, 7, 0), is the ray AB, with A = (5, 3, 0) and B = (4, 6, 0). is the segment Fp, with P s (2, 1, 0) and Q = (0, -7. 0). is the segment RS, with R = (2, I, 1) and S = £0, —1,1}, 12.2 A DISTANCE FORMULA In this section we develop a formula for the distance between two points expressed in terms of their coordinates. We begin by consider- ing an example. Example 1 Let A - (3, 2 t 1) and B = (5, 4, 2). (See Figure 12-5.) I .el A a = (3, 2, 0), B t = (5, 4, 1), B 2 = (5, 4, 0), B 3 = (0, 4, 1), and B* = {0. 4, 2). Then AA2B2B1 is a rectangle and ABi = Ajfiz. Also BBiBzB* is a rectangle and B t B = BsB 4 , Bjl Figure 1S-5 To find A2B2 we use the Distance Formula and the x- and ^-coordinates of A 2 and 13* In the xy-eoordinalc system. As = (3, 2) and B 2 = (5, 4). Therefore A 2 B 2 = V(5 - 3)* + (4 - 2) a = \/8 = 2V2. 520 Coordinates in Space Chapter 12 To find B3B4 we use the Distance Formula and the y- and subordinates of B3 and B+ In the ^-coordinate Systran B 3 = (4, 1) and B 4 = (4, 2). Therefore B S B^= V(4 - 4)z + (1 - 2)2 = VT= L Now BBi is perpendicular to the xy-pkne. It is also perpendicular to every plane parallel to the ay-plane. In particular, S?i is perpendicular to the plane with equation !■! Then mh & perpendicular to every * — * ■ — * line in that plane through tfj. Therefore BB t _L AB t and AABiB is a right triangle. It follows from the Pythagorean Theorem that {AB)* a (AB l )2 + (B^)* (AS)* = (A 2 i3 2 ) 2 + (£ 3 B-i) 2 (AB)* = (V§) 2 + I 2 (AB)S = 8+1 = 9 AJ3 s 3 Following the procedure used in this example we shall prove the next theorem. THEOREM 12.1 {Distance Formula Theorem) The distance between F{x lt y it z{) and Q{x& y% z 2 ) is given by PQ = Vta - *i)» + (sf! - yir + (* - *)*. jftoo/* Let P(*i, y lp si) and ()(x a , 1/2, afe) be given. Let P x = (xi, yi f 0) Qi = (*fe gfe 21) Qz = (*a, y 2 , 0) <?a = (0, y 2 , xi) Figure 12-6 shows these seven points as distinct points. Depending on die values of the coordinates, the points may turn out not to be distinct. For example, if zj = 0, then P = P x . We proceed to find PQ in terms of the coordinates of P and Q. If F and Q\ are distinct points, then they lie on a plane parallel to the u/-plane and PQ ± is parallel to the xy-plane. Therefore (1) PiQtQiP Is a rectangle, or (2) P = Ft and ft = Q 2t or (3) F = Q L and P, = £ z . 12.2 A Distance Formula 521 Figure 12-0 In all three cases it is true that (FQ 1 f = ftQtf = fa - *i)» + (y* - yi) 2 - Similarly, (I) QQ1Q3Q4 is a rectangle, or (2) Q = Q 4 and £i = Qz, or (3) Q = pi and p 4 ■ Qn In all three cases it is true that If 9 = p Xi then % = s 1? (sz — zi) 2 = 0, and Hf = (PQtf = {x 2 - xj* + (tfz - y,) 2 + (z 2 - ft) 2 . If P = p l5 then 12 = afe $2 = 5fo (*2 - *i) 2 = 0. (1/2 - gfo? = ft and (FQ)2 = (p!p) 2 = (ss - *i) 2 + (t/2 - f/0 2 + (*b - *i) 2 - If Q ^ p x and P 7*= pi, then ^5, 1 FQi> mid it follows from the Pythagorean Theorem tliat (Pp) 2 = (FQtf + (WrP, (Pp)2 = fe - X i)» + (y 2 - yi) 2 + fo - s,} 2 . We have now exhausted all cases. For if P and Q are any points and if pi is related to P and Q as indicated in the first sentence of this proof, thenP = Pi; or Q = Qn or P, Q, Q\ are three distinct points. We have proved in all of these cases that (Pp)2 - fa _ Xl ) 2 + (tfz - yi) 2 + (^ - *i) 2 - Therefore pq = vte - xi) 2 + (ys - yi? + te - »i)*"- 522 Coordinates in Space Chnptsr 12 EXERCISES 12,2 ■ In Exercises 1-5, find the distance between Panel Q. LPs (0, 0, 0), Q = (3, 4, 5} 2. F=(2, -1.7), Q= (2, 1.-7) 3. P=(7, 8,1), p = (0 t fl,0) 4.P={0,0,5),@ = (3,4,0) 5. P=(a -J,5),(? = (1, -1,6) 6* Use the Distance Formula to show that die following points are collinear: A m (-1,5, 2), B = (5, 7, 10), C = (2, 6, 6), 7. Use the Distance Formula to show tliat the following points are not collinear: A = (0, 0, {)), B = {5, 3, -4}, C = (15, 10, - 12). a If A = (L 3, -2), H = (2, 7, 1), C as (2, 6,0), is A ABC a right triangle? 9. IfA = (5,3,2), B a (7,8, 1), C = (5,2, -3), is AABCa righttriau I ' 10. Find the perimeter of &DEP if D = (0, % 0), E = (0. 6, 8). F = (10, 6, 8). ■ In Exercises 11-14, find tl*e distance between F{2. — 1, 7) and the given plane a. 11. a = ((x, y. s) : x = -5} 13. a = {(x, y, =) : z = 0} IS. a={{x,y,z):y = 0) 14. a : = {(x, y, z) : y = ~1) ■ In Exercises 15-18, find the distance between the line / given by I = {(x, y r z):x = 3, y = 4} mid the given plane a which is parallel to it. (The distance between u line and a plane parallel to it is not del ined in the book. 'Write a suitable definition for tliis distance.) 15, a = {(r } y t z) : x = 0} 17. a = [(«, y, z) i ; ij a 6} 16. a — {(x, y s z) : i = —5) 18. a m {(x, y, s) : tj = 4} ■ In Exercises 19-23, sketch a figure to represent the given set. (The graphs will be three-dimensional.} 19. A = {(*, y, s) : 1 < * < fi, £ < y < 4, < ;= < 10} 20. B = {(x, y, 5) ; 1 < a: < 2, 6 < y < 8. 4 < 2 < 6} 21. C = {(x,tj,z):l<x< 2, 10 < y < 12, < = < 10} 22. D = {{x, y> z) : 1 < x < 2, 14 <, y < 16, 4 < z < 5 or 6 < a < 7) S3. £ = {(*, y, sj : 1 < x £ 2 t 18 < «/ < 20, < z < 10} 24, challenge problem. On one large sheet graph the union of the sets in Exercises 19-23. 12.3 Parametric Equations for a Una in Space 523 12.3 PARAMETRIC EQUATIONS FOR A LINE IN SPACE Before studying Section 12,3 you should review the work on para- metric equations in Chapter 11 . By use of a proof similar to the one for Theorem 1 1.7 the following theorem can be established, A proof of this theorem is not included here. THEOREM 12.2 If P(* 1t yu -i) and Qix 2 , ua, z 2 ) are any two dis- tinct points, then x = xi + k(x z - x\) t W$ = (x, y, z) : y = tfi + k(tf2 - yi)> and k is real z = si ■+■ kizz — «i), If R a {a, b, c) t where a = x-i + k{xz — x\), h= yi + %j - tji), c = zi + k(2t - *i), is a point of FQ t then It is the point P R£PQ and PR = k * Pp H € <W P£ and PR = -Jfc ■ P@ Example 1 Given P = (0, 0, 0) and Q = (3, 1, -2), find the point R on P$ such that PR = 3 ■ FQ and the point S on opp PQ such that Solution; P$ = {(x, U, z) : x = 3fc, y = *, * = -2fc, fc is real} To get fl % set fc = 3. Then R = (9, 3, -6). To get 5, set k = —3. Then S = (-9, -3, 6). Bampfe 2 Given A = (5, 1, -2) and £ s (7, 3, -2), find the mid- point M of AB. Solution; £jf = {(*, y, s) : x = 5 + 2fc, y = 1 + 2*, 2 = -2,kis real}. The ^-coordinate of A is 0; the fc-eoordiuate of B is 1 ; of M is £. M = (5 + 2 ..^ 1 + 2 - J. -2) = (6, 2, -2). if k = 0, tf Jt>0, if k<0. 524 Coordinates in Spact Chapter 1 2 Example 3 Given £ = (5. 7, 9} and F s (— 2, 3, —4), write para- metric equations for Et. Solution: x - 5 - 7Jt y = 7 -4k s = 9 - m *£ Example 4 Given 11 = (5, -3, 14) and J = (-2, 5, - 1), write para- metric equations for lit (1) with a parameter Jfc so that A: = at H and & = 1. at 1. (2) with a parameter h so that /« = at I and fa = 1 at H. (3) with a parameter t so that f = at 7/ and t = 10 at L Solution: (1) x= 5 - 7* (2) * = -2 + 7ft (3) x = 5 - 0.7* y = -3 + Bk u = 5 - 8/i y = -3 + 0.8* x= 14-15* S=-1 + I6ft z= 14-1.51 G < fc < 1 < « < 1 < f < 10 Example 5 Given J = (5,1, 3) and K s (7, —1, 5). find an equation for the perpendicular bisecting plane of JK. A simplified equation is desired. Solution* Let a be the perpendicular bisecting plane of JR. Then P(x> y, s) is a point of a if and only if JP a EK, tff)» - (IX)*, (* - 5)2 + (y - 7)2 + (a - 3)2 = (x - 7) 2 + (y + If + (s - 5)2, x- 2 - lOx + 25 + y« - 14?/ + 49 + s* - 6s 4- 9 = x 2 - I4x + 49 + y 2 + 2y + 1 + s 2 - 10s + 25, - lOx + 14x - Uy - % - 6s + 10s + 25 + 49 + 9 - 49 - 1 - 25 = 0, it - 16y + 4z + 8 = 0, x — 4ij -r z + 2 = 0. Then x — Ay + s + 2 = is "an equation of a" This means that a = {(x, y, s) : x - 4y + s + 2 = 0}. 12.3 Pwsroetrk Equations for a Line in Space 525 EXERCISES 12.3 In Exercises 1-4, find the coordinates of the point P satisfying the given condition. L A = (0 T 0, 0), B = (3, 3, 3). P€ 3$. AP = §>AB, 2. A = (0,0,0>,B = (3,3,3),P€ W a5,AP = * 3 *AB 3. A = (5, -2, 6). £ = (1, -3, 2), P C HA» AP = 2 ■ AD. 4. A = (-3, 2, 5), fl = (5, 0. - 1), P € XB t AP = FB. In Exercises 5-10, A = (0, 0, 0), B s (6, -3, 12), C = (-9, 18, 9). 5. Find D, the midpoint of J3fl 6. Find B, the midpoint of ACT, 7. Find F, the midpoint of A B. 8. Find Gi, the point that is two-thirds of the way from A to D, 9. Find G2, the point that is two-thirds of the way from B to E. 10. Find G& the point that is two-thirds of the way from C to F. 11. Given S = {(x, y , z) : 3x + 4i/ + 5s = 50), which of the following points are elements of S: A(5 t 5, 3), fl(G, 0, 5), C{- 1, -2, 12}? 12. challenge problem. Find tlie point on the T-iutis that is equidistant from A{2, 3, 1) and B{7, 2, 1). 13. challenge problem. If S is the set of all points each of which is at a distance of 7 omits from the origin. End an equation (in simplified form) for S. What is a set of points like S usually called? 14. challenge pboblem. If T is the set of all points, each of which is at a distance of 3 units from the point (3, —1.2), find an equation (in simpli- fied form) for T. 1.5. CHALLENGE PROBLEM. (Jiveil S = {(-Y, y, a) : & + f 4- z 2 = 125}, T = {(x, y, z):x = 5}, mid Q = (5, 0, 0), prove that all points P in the intersection of S and T are at the same dis- tance from Q. 16. challenge problem. Let planes u and and points, O t A, A Ir B, B^ be given as follows: /? = {(*. y, z) : j 4- y + s = 9}, O = {0, 0, 0), A = (2. 4.2). Ai = (6.0, 0), B = (3, 3, 3). Bi = (9. 0, 0). (a) Show that A and A i are points; of a. (b) Show that B and Bi arc points of ft. 526 Coordinates « n Space Chapter 12 :V) Show that UA 1 ZKl. (d) Show that UB 1 SSI, (e) Find a point Aa in a such that A, A], A* are noneollinear. (i) Find a point H* in /? s uch llial B» B lt Ba are non colli near. (g) Show that UA ± AAg, (h) Show that US 1 SSi. (!) Show that DA 1 a. (j) Show that DZJ 1 j8. (k) SIiow that 0, .4, iB are colHnear. (1) Find the distance between a and /i. 17. CHALLiiscjK i-itonr.nM. I.et planes a and /3 and line / lie given as follows: a = {(x t y, z) : x + y + 2,? = 8}, = {(*, y, s) : x + y + 2* = 2}. = [{x, i/, z) : x = y = ^H (a) Find the point A in which / intersects a. (b) Find the point B in which / intersects /?. (c) Prove that A~B 1 a. (d) Prove that AB _ fi, (e) Find tlie distance between a and /?. 12.4 EQUATIONS OF PLANES In Chapter II we used coordinates to write equations of lines. For example, 3* + Atj +■ 7 = is an equation of a line in an ty-plane* We say that 3x + 4y + 7 = is a linear equation in y and t/. The equation ax 4- ^ 4- G = is called the general linear equation in x and y. It is natural to extend this terminology and to call ax + hy -h cz 4- & — the general linear equation in x, !/, z. In this section we show that, if an xyz- coordinate system has been set up, every plane has an equation of the form ax + by + '* + d — 12.4 Equations of Plane* 527 with «, h, & d real numbers and with a t b, c not all zero, and conversely, that every equation of this form is an equation of a plane. What is the situation ifa,h t and c are all zero? More specifically, what is the graph of Qat + % + G*+I = 0? Of Ox + Qy + CMr + = 0? We begin with several examples, assuming that an xy ^-coordinate system has l*ecn set up in space. Example t IaH = (0, r 0), P = (3, 4. -6), and let a be the plane that is perpendicular to UP at P. We shall exjyrefa a in terms of the co- ordinates of the points on it using set-builder notation, Now OP = (fr tf % z):x = 3k>y = 4*. z = -6*, k > 0}, I ei: O I h 1 1 1 1 point on OP that is twice as far from O as P is from O, To find the coordinates of Q, set k = 2, then (J = (6, 8, - 12), P is the midpoint of UQ t and a is the perpendicular bisecting plane of U%). The perpendicular bisecting plane of a segment is the set of all points equidistant from the endpoints of the segment. Using the Dis- tance Formula, we find that the distance between O(0, 0, 0) and (x, y, z) is V* 2 + yf + a? and that the distance between P{6, S, — 12) and (x t y t z) is V(* - 6)2 -f- {y - 8)2 + ( S + 12)3. Therefore « = {(X, y, Z) : # + 7 + ^ = v'«* - 8)* + {y - 8) 2 + (* + 12) 2 }. a = {(r, p, c) : s* + y 2 + s 2 = x 2 - 12% + 36 + t/2 - 101/ + 64 + z 2 + Uz + 144) f « = {(*, y. s) : Hx + 1% - 24s - 244 = 0}, « = {(*. y» s) : 3* + 4y - 6* - 61 = 0). Example 2 Let O = (0, 0, 0% P = (3, 4, - 6), and let /? be the plane that is perpendicular to OF at O. Note that /? is parallel to the plane a of Example 1 and that a. docs not contain O, but that fi does. Now OP = ((*, r/ s z):x = 3k, y = 4Jt, z = -6k, k Is real}. Let R be the point on opp OP such that OH = OP To find the coordi- nates of R, set k = — 1, then K sb (—3, — 4, 6), and O is the midpoint of BP and $ Is the perpendicular bisecting plane of KP. 528 Coordinates in Spac* Chapter 12 /? is the set of all point 4 ! (x, y, z) each of which is at the same distance from (3, 4, -6) as it is from (-3, -4, 6), Therefore P = (fe .'/> A = V(« - 3? + (y - 4)» + fr + 6)2 = ^ + 3)> + (y + 4)* + («-6)S} t fi = {(*, y, «) ■ : x s - &x + 9 + i/2 - 8y + 16 4 z 2 4 12s +- 36 = x 2 + 6* 4- 9 + !/ 2 + 8y + 16 + s£ - 12s 4 36}, p a ((* //, 4 : -12* - 16y + 24* = 0}, = {(*, t h Z) ; 3* 4 4y - 62 = 0}. ficample 3 Let y he the following set of points in .fys-space: y = {(*, y, *) : * - 2y + 3a-4 = 0}. We shall show that 7 is a plane, Xote in Example 1 that an equation of a is 3* + 4y - Qs - 6] = T that the coefficients of x, y, z in this equation are 3, 4, — 6, that the foot of the perpendicular from O to a is {3, 4, —6), that 32 + 4 s + (_e)* - 61, and that 61 is the negative of the constant term in the equation of a. Tliis suggests that we transform the given equation of y, x - 2y 4 32 - 4 = 0, into an equivalent equation in which the sum of the squares of the co- efficients is the negative of die constant term, that is, an equation ax + hy + cz + d = where a 2 4- b' £ 4 e 2 = —d, Wc multiply through by k and then determine a positive value of k so that the resulting equation of y has the desired property. We get fcr - 2ky 4 3Jb - 4fc = and we want fe3 + ( _2fc)* 4 (3fc) 2 = 4* with fe > 0. Therefore # + 4/t 2 + 9ft* = 4k, 14*2 = 4^ fc m 2 12.4 Equations of Planas 529 Therefore Note that and that $ is the negative of the constant term of our **adj listed" equa- tion of v. Taking a clue from Example 1, we suspect that y is the plane that is perpendicular at (7,-7,7) to the line through (0,0 4 0) and {f t — * I }, and hence that it is the perpendicular bisecting plane of the segment with endpoints (0, 0, 0) and (4, — J, 4^ ). To show that it is, we let y' be that perpendicular bisecting plane and we show thai y = y> Now y ' is the set of all points (st, y, z) each of which is the same dis- tance from (0, 0, 0) as it is from (^, — ■$» 4^- ). Therefore Y m {{x, y, z) : V* 2 + U 2 + ^ = V(* -*) 1 +(* + *)■ + <»- V M. / = ft*. £f, «) : x 2 + if + ~ 2 Therefore y' = y t and y is a plane. Since y = [{*, r/, s) : x - 2y + 3s - 4 = 0}, this proves that x - 2y + 3z - 4 s is an equation of a plane. Example 4 Let 5 be the following set: 5 = {fr y,z):x-2y + 3z = Q). Since (0, 0, 0) is an element of 5, it follows that S is a set of points that contains the Origin. We shall show that 5 is a plane. Taking a clue from Example 2, we suspect that 5 is the plane that is perpendicular at the origin to the line through the origin and (1, — 2, 3). We suspect further that 8 is the perpendicular bisecting plane of the segment with endpoints (I, —2, 3) and (— 1, 2, —3). Let 8' be that perpendicular bisecting plane. Wc shall show thai 8' = 5, 530 Coordinates In Spae* Chapter 12 We have 5' = {(x, y, z) : Vfr - l) 2 + {y 4- 2)* 4 (z - 3) 2 = V(* + I) 1 + (y - 2J 2 + ( Z + 3)*}, 6" = (fr y t z) : x 2 - 2* + 1 + tp + Ay + 4 + z £ - 6z + 9 = x 2 + 2r + I + y 2 - 4y + 4 + s 2 + 6s + 9}, ^ = {(a:, y, z) ; -4* -f 8y - 12z = 0}, 5' = {{x > y > s):x-2 !/ + 3* = Q}. Therefore $* = 8, and 8 tea. plane. We arc now ready for several theorems on linear equations in £, y t and z, THEOREM 12,3 Given an aryz-eoordinate system, every plane has a linear equation. Proof: Let a plane « !>e given. We consider two cases. Case I. O(0, T 0) does not lie on a. Case 2. 0(0, 0, 0) lies on <*. Proof of Case 1: Let P{a, b t c) be the foot of the perpendicular from O to a. Then a, h, c are not all zero. Let Q = {2a, 2b : 2c). Then P is the midpoint of UQ, and a: is the perpendicular bisecting pkne of UQ, Then a = {(*% y, z) : \/x* + y 2 + z 2 = V(* - 2») 2 4- (y - 2£>} 2 + (z - 2c:) 2 }, 0£ = {(%, y, z) : ax + by + cz = a 2 + b 2 + c 2 }. Therefore in Case I, a has a linear equation. Proof of Case 2: Let P{a, b, c) be a point distinct from O on the tine perpendicular to a at O. Then a t h, c are not all zero. Let Q = {—a. —b, — c). Then O is the midpoint of PQ and q is the per- pendicular bisecting plane of FQ. Then «= {fcy,*) : Vfr - a) 2 + (y - *>) 2 + {z - c) 2 = >/(* + «) 2 + fr + *>) 2 + (* + «)"), a = {(%, y, *) : x 2 - 2a* + a 2 + y 2 - 2fci/ + fe 2 4- z 2 - 2zc + e 2 = jc 2 + lax + a 2 4- y 2 + 2by + fe 2 + s 2 + 2sc + c 2 }, a m {{x, y, z) : ax 4- by + cz = 0}. Therefore iu Case 2, a has a linear equation. 12.4 Equations of Plum 531 THEOREM 12.4 Given an ^-coordinate system, the graph of every linear equation ax 4- by 4- ez + d = 0, in which a f b t c,d are real numbers and a., b, c are not all zero, is a plane. Proof: Let an sy^-coordinate system and an equation ax -J- % + cz + rf = 0, with a, A», c not all zero, be given. We consider two cases. Case J. <*#(). Case 2, d = 0. Proof of Case 1 ; Suppose d =£ and let « = {(x, y, z) : ax + % -f cs + rf = 0}. Let /: l)e a number, not zero, to be specified later. Then a = {(x, y T z):a f x + b'y -f c'z + d' = 0} : where a 1 = fa b* = kh, u' = kc, and ff = hd. Taking a clue from Example 3, wc shall determine k t with Jt ^ 0, so that d' = -(a'}2 - gtp - {&)*. Substituting we get kd= -{kaf -{kh)* - (kc) 2 . Solving for k we gel kd = -Wo 2 - Ffe2 - k 2 c\ d = -ka 2 -kb* -kc 2 , d = k(-a2 - fea _ ^ and this is our specified value of h Note that since a, b t c are not all zero, then at least one of the nonnegativc numbers, a 2 , h 2 , c 2 , must be positive, and a 2 4- h 2 + c 2 is positive. Hence the expression for k in terms of a, b, c, cJ is mathematically acceptable since the indicated di- vision is division by a number that is not zero* Checking wc find that tt = kd = #(-a2 - 68 - c2) = — (Jfcu) 2 - (Mi) 2 - (fee)*, d = -(*>)* - (by - { C y. 532 Coordinates in Space Chapter 1Z Let P = (a*, b\ c 1 ) an d Q = (2a' f Zb\ Icf). Let cr' be the perpen- dicular bisecting plane of OQ. Then or' = {(x, y, 2) : V* 2 + y 2 + £ = >/<* - 2a*)* + ( y _ 2bT + > - 2c 1 ) 1 }. = {*, i/, s) : *a + y* + & = * 2 - 4rf* 4 4(a')2 4 y 2 - 4b'y 4 4{fc') 2 + # - 4c'.s + 4(c') a }, = (fo y, z) ■ u'x + //./ + <te - («7 - (PT - WT - 0), = {(*, y, s) : a'* + fc'y 4 c's 4 tf - 0), a {{x, y, z) : kax + kby 4 fcos + kd — $}> = {{ar, y, z) : ax 4- by + cz + d = 0}, and a' = a. Since & t h a plane, it follows that a is a plane, as we wished to prove. Proof of Cose 2; Suppose that r / = and let a = {{x t y, s) : ox 4 by 4 cs = 0}, Then a = {(x, y, z) : ox 4 by 4 cz 4 ii = 0}. Let P s (a. 6, c)„ p = (— tt, — ft, — c), and let a' be the perpendicular bisecting plane of TQ. We shall show that a' = a, and this will prove that ax + by + cz + d = is an equation of a plane. Now «' = {fa y, z) \ Vfr - tt?» + (y - fr) a + fr - cp = \/(x 4 u) 2 4 by 4 b)'* 4 {* 4 c) 2 }, a* = {(*, y } *) : a 2 - 2a* 4 0* +• y 2 - 2% 4 P 4 z 2 - ^SJ 4 c 2 s x' 1 4 2a* 4 a 2 + y t + 2/?y 4 b 2 4 z 2 4 2zc 4 c 2 }, a' = {(x, y, z) : —4ax — 4by — 4cz = 0}, ex s {{r, */ t 2) ; ax 4 % 4 ra = 0} , "Therefore a' — a. THEOREM 12,5 If a, b, c, J are real numbers with a, b, c not all zero, then the plane a = (£x, y, z) : ax 4 by 4 cs 4 d = 0} is perpendicular to the line through 0(0, 0, 0) and P{a, b, c). 12.4 Equations of Plan** 533 Proof: If tt = 0, then as in the proof of Case 2 of Theorem 12.4 it follows that o is perpendicular to the line Py y where Q = (—a, —h, —c) and P = (a, b t c). But O is the midpoint of PQ. Therefore Op = P<$ and a 1 OP. If d t^ 0, then as in the proof of Case 1 of Theorem 12.4 S it follows that a is perpendicular to the line OQ, where Q = (2a\2b\2t/)> Saw 0<$ = {{*, tj, z) : x = 2aft, y = 2b% z = toft, t is real}. (We have used t as the parameter since k was used for another purpose in the proof of Theorem 12.4*) Setting * = -i, we obtain a particular 2k point on OQ, namely (|- , -|- , ^~) = P(a, b t c). Since a is perpendicular K ft 1\. to Op, and since 5iy = OP, it follows that a is perpendicular to 5?. Example 5 Find an equation of the plane perpendicular to OP at F if O = (0, 0, 0) and P = (2 4 -7, 0). Solution; For every real number d, 2x — ly + d = is an equation of a plane perpendicular to OP. (Which theorem or theorems of this section are we using here?) To get a satisfactory equation we fix d so tlial the plane contains P. Substituting the coordinates of P, we get 2-2-7(-7) +d = and d = -53, Therefore 2x - 7y - 53 = is an equation of the plane perpendicular to Or at P. Example ft Find an equation of the plane perpendicular to DP at O if = (0, 0, 0) and P = (2, -7, 0), Solution: For every real number d, 2x — 7tj + d = is an equation of a plane perpendicular to OP. The origin O(0, 0, 0} is a point of this plane if2*0 — 7*0 + 4 = 0, that is, if d = 0. Therefore 2x — 7y = is an equation of the plane perpendicular to UP at O. 534 Coordinate* in Spact Chapter 12 Example 7 Find an equation of the plane perpendicular to OP at P if O = (0, 0, 0) and P = (7, 4, -5). Solution: 7* + 4y - 5z + d = 7.7 + 4.4 _ 5(_5) + da 90 + d = d a -90 7* + Ay - 5s - 90 = Example 8 Given o = {(*, t/, 2) ; 5sc - 2y + 4a - 8 = 0}, find u point P .such that « is perpendicular to the line through O(0, 0, 0) and P. Solution: P - (5, -2, 4). Example .9 Given a = {(x t y, *) ; 2k - 7* = 0}, find a point P such that a is perpendicular to the line through O(0, 0, 0) and P. SbFtfftbn; F = (2 f 0, —7), Example W Find an equation of the plane a that contains the points O(0, 0, 0), A(l, 1, 1), and B(l t 7, -3), Sofuft'cm; We want real numbers a, b, c y d with a, b, c not all zero such that a = {(x, y, z) \ ax + by + cz + d = 0). Since O, A, B arc points of a, it must he true that (1) a*O + ft*O + 0*O + <la>0, (2) a •! + b'l + cl + d = (), (3) a • I + b • 7 + &•{— 3) + 4 = a From (1) we deduce that d = 0. From (2) and (3) we deduce by sub- traction that 1 -6/;-f4c = 0. 12.4 Equations of Plan** 535 Now neither b nor c can be zero. Why? To complete the solution we take any number except for b and solve for the corresponding a and c. Thus, if b = — 2, then c = — 3, a = 5, and a a {(x, y, z) : 5x - 2y - 3* = 0}. To check, substitute coordinates as follows: O: 5*0-2-0-3-0 = 0-0-0 = 0. A: 5-1-2-1-3-1 = 5-2-3 = 0. B: 5*1-2-7-3- (-3) = 5 - 14 + 9 = 0. Kxamplc II Find an equation of the plane that contains the three points P(l, 5, 7), Q(-l, 2, -4), R% 1, -5). SoJti (ton; (1) ax + % + ez + d = (2) a • 1 + /? - 5 + c • 7 + a" = 0, a + 5b -f 1c + d = (3) fl-(-l) + ft-2 + c-{-4) + d = 0, -a + 2fo-4e + d = (4) a -2 + fc * 1 + c - (-5) + 4 = 0.2<t + &-5c+<Z=0 From (2) and (3) we get (5), and from (3) and (4) we get (6). (">;> 2a + 3& + lie = (6) -Za f b + c = From (6) we get (7), and from (7) and (5) wc get (8). (7) -9a + 3b + 3c = (8) llo + 8c = Take a = 8; then c = — 11. From (6) we get fc = 35 and from (2) we get d = — 106, A satisfactory answer is 8* + 35y - llz - 106 = 0, Check: P: 8-14-35-5-11*7- = 8 + 175 - 77 -- 106 = 0. Q: 8-(-l) + 35-2 - 11 -(-4) - 106 = -8 + 70 + 44 - 106 = 0. R: S-2 + 35-1 - ll-(-5)- 108 = 16 + 35 + 55 - 106 = 0. 536 Coordinates In Span Chapter 12 THEOREM 12.& Consider the plane a ={(*» if* x ) : ax + % + <* + d - °)' If « = 0, a is parallel to the x-axis. If h = 0, a is parallel to the y-axis. If c = 0, or is paralU-] to the --axis. ftoo/v Suppose first that a = and d = 0. Then a = {(*, y t z):by + cz = 0} and for every real number x, a contains the point (jr., 0, 0). Therefore of contains die ar-axis and is parallel to it. Suppose next that a = and fi 7^ 0. Then and since a = {{x, y, z) : by + cz 4- d = 0} b-O + c-O + d^O, it follows that no point (x, 0, 0) is a point of a and henec that the x-axis is parallel to id Similarly, it may be shown that if b = 0, men a is parallel to the ff-axis, and that if C = 0, then a is parallel to the s-axis. Examph 12 Let a, /?, y, 5 be given as follows: a = {(*, y t z) : 3* + 4y - 12 = 0}, /?={(*, i,,z):3x-2z = 0}, 7 = {(x, tj, z) : 5tj = 8}, * = {(*, i/, «) : 2* - 5 = 0}. (See Figure 12-7,) & is parallel to the s-axis, a contains the line I in the xy-planc widi x- and (/-intercepts 4 and 3, respectively, and every line parallel to the s-axis that passes through a point of I Figure li-7 12.4 Equations of Plane* 937 (See Figure 12-8.) /? is parallel to the y-axis; in fact, it contains the i/-axis. The plane /? contains the line I in the .re-plane that passes through the points ((I } 0) and (2, 0, 3). It contains every line parallel to the y-axis that passes through a point o( L Figure 12-8 (See Figure 12-9.) y is a plane parallel to the ar-axis and the 3-axis; hence it is parallel to the .ts-plane. y contains every point whose (/-coor- dinate is 1.6 and every point of y has y-coordinale 1.6. ♦a yml(x,y,i};5y>B\ Fignn- 12-W (iSee Figure 12-10.) 5 is a plane parallel to the y-axis and to the >axis; hence it is parallel to the y£-plane. The plane 5 is the set of all points whose ^-coordinate is —2.5, *- -if*, y.*): 2* + *-0f Figure 12-10 538 Coordinates in Space Cnap:e'12 EXERCISES 12,4 ■ Let a plane « = {(*, y, s) : 4x — 3y + 180 = 0} be given. In Exercises 1-10, determine if the given point is a point of a, 1. (0, 90, 0) 6. (0, W, 137) 2. (45.0,0) 7, (-45, 0,237) 3. (0,0,180) 8, (-30.20,10) 4. (0, 0, 0) 9, (7, 69, j) 5. (0, 0, 37) 10, (-100, -73± 6Xif) B In Exercises 1 1 -20, an equation of 8 plane is given. Find three numbers Or b, c f not all zero, such that the given plane is perpendicular to the line through the origin and P{a, b, c). 11. -3.r - 7y + 14 = 16, f-r ? -nl=0 12. 2x - y + z - 73 = 17. x - y + 10s = 13. 2y = 3x 18. x - y + 10s + 10 = 14. tj = 3x + 5 19. y + lOx - 4s = 15. * = 3y — 3 20. x - y = s - 1 ■ In Exercises 21-25, siin plify the given equation to an equivalent equation in the form of the general linear equation ax + by + ex -+■ d = 0. If one of the coefficients is zero, Uiat term may be omitted in the simplified form. Thus we accept 3* + 4y = in place of 3* + 4tj + Oz + = 0. 21. yx" + tf + z* = V(x -l )»f (y-2)' + (g- 3) s 22. y '{* - 2}* + (y - 3?' + s' = \ /^T 2f+(y + 3)' + ^ 23. V fx- I) 2 + (r/ - if ^UP e V (x- 12j a + ( y - 5) a -f s* M, y ^T4) g + (y + 4)» + (a -T ffi = V* 2 ^ y» + ^ 25. Vfx^T? + ~y*" + [5 +TP = y/{x - 1)* + (y - 1)2 + (z + 1)2 ■ In Exercises 26-30, find a linear equation in simplified form for the plane that con tarns die diree given points. 26. (0, 0, 0), (8, a 0), (-3, 0, 1) 29. (8, 0, -4), (0, 4, 3), (4, 4, 1) 21 (1, 3. 1), (0, 1, 1), (2, 1, 1) 30. (3, 0, 1), (-9, 0, -3), 23, (5, 2, 3), (-3. 6. 7), (0, 4, 6) (51, 10, 17) B In Exercises 31-40, ail equation of a plane is given. Find the coordinates of its intercepts on the coordinate axes. If it does not intersect an axis, write "none." The answer to Exercise 31 is given as a sample. 31. 2x - 3s + 30 = x-ratercept; (-15, 0, 0) {/-intercept: None s- intercept: (0. 0, 10) 12,5 Symmetric Equations tor a Un« 539 32. 3x + by - 2x - 20 = 37. tj = -3 33. * - y + s = 38. z = -4 34. 5* - 4y + 100 = 39.x + y -J- e = 100 35. 2x - ij + -s = 40. a : 4* y + = + 100 = 36. x = 6 In Exercises 41-50, state whether ihc given set S is a plane, the union of two distinct planes, n line, the union of two distinct lines, a set consisting of a single point, or the null set. 41. S = {(x, y t z) : x = 2, y = 3, z a 4} 42. S = {(x, y. a) : x + y + s = 5J 43. S={{x t ,j,z):x+,j = 5} 44.S={(x t y):x+y = 5) 45. S = {(», y) : x = 5} 46. S = ((x, y, z):x = 5orx= 10} 47. S = {(x t y, x) : x + p + « = 10 and x + y + z = 20} 45. S = {(x, y, x) : x + y -h z= 10 or x 4- y -f s = 20} 49. S = {(x, y, z) : x + y + 2* = 10 and x - 2y + 2z = 10} 30. S = {(x, i/, c) : x + y + 2s = 10, x - y + 2* = 10, 2 = 0} 12.5 SYMMETRIC EQUATIONS FOR A LINE Since a linear equation in x and y represents a line in an xry-plane and a linear equation in x, y> z represents a plans in xj/z-spacc, it is nat- ural lo ask how we might represent a line in terms of coordinates in xys-spaee and to do it without using a parameter. A clue to the answer is provided by some of the exercises of Exercises 12.4. As usual, we assume an xyx-eoordinate system is given. Example 1 If P = (2, l s 7) and Q = (5, -1, -3), then The notation x- 2 y- 1 x- 7 5-2~-l-l : -3-7 is a short way of saying that ill, y 1 and y 1 = *- 7 . 5_2 -1-1 -1-1 -3-7 540 Coordinates in Space Chapter 12 To see that the representation in terms of these equations is correct, note that * - 2 - y - J rt „i j/- 1 _ ~- 7 unci 5-2 -1-1 -1 - 1 -.3-7 arc equations of distinct planes, one of them parallel to the s-axis and the other parallel to the it-axis. Note that both F and Q lie on each of these planes. (Check this by substituting coordinates.) Hence FQ is precisely the intersection of these two planes and the representation we gave in terms of "symmetric" equations is a satisfactoiy one. Actu- ally, there is more symmetry in this situation than we have indicated so far. It is natural to think of ry as die intersection of three planes related to the three expressions equated to each other in the set-builder notation, Thus every point of FQ lies on each of the following three planes, a t /?, and f, which are parallel, respectively, to the a>, y-, and 2>axes: * = [<* * *> ■■ fff = -^nrr)- Generali zing the situation in Example 1, we obtain the following theorem. THEOREM 12. 7 If P(x lt y it *i) and Qfa y 2i z 2 ) are two distinct points with x% ^= 4*2, tj\ ^= tj>£, £\ ^ jgtg, then FQ = { lx t y, z) : = -* *— = - I X2 - Xl yz - Jfr 8£ - «i ftw/: Let I »2 — 3^1 ^2 — 2»i ) I x 2 — xi y 2 — yi J 12.5 Symmetric Equations for a Lin* 541 Now P(xu j/i, 2i) is a point of a since — ^- = — — = y* — yi z z ~~ *h and Q(x 2 , ijt, %) is a point of a since - /a " yi = * 2 ~ Zl = 1. ^a - 1/1 «■— »i Similarly, we can show that P and p each lie on both /3 and y , Also, a, /i, y are three distinct planes. For a contains the point (xi — X t tf], Zi} t whereas ft and 7 do not; fi contains (xi, 1/1 — 1, Zi), whereas on and y do not: y contains (xi, yi,Z\ — 1), whereas a and fi do not The — 1 here may appear to border on the magic. Actually it does not For example, if Xo is any nonzero number whatsoever, then a con- tains (xi — Xo t yi, zi), whereas fi and y do not Therefore (x, y, z) is a point of PQ if and only if it is a point of all three of the planes a, fi, y, and hence the representation using sym- metric equations is a valid one. THEOREM 12.8 If F(x u 1/1, *t) and Q(x 2 , y 2 , $§} arc distinct points with x-j ^= x 2 and y^ sfi y 2 > then I X 2 - Xl $2- IJl > Proof; X — Xi u — 5/1 , and z = «i ia - xi ya - ^3 are equations of two distinct planes each containing both P and Q, Therefore PQ is the intersection of two planes, and the representation given in the statement of the theorem is a valid one. Similar proofs may be given for Theorems 12*9 and 12,10. THEOREM 12.9 If F(x u yu z t ) and p(x 2 , yu z^} are distinct points with xi ^= x 2 and z± =£ ffg, then i v*> y> « X — Xi * — Zl '} Zi \ X 2 — X| Z% — Zi i 1 HEO /t EM 12.10 Tf F(Xi , «/ 1, ?i ) and ()(xi , i/ 2 , z 2 ) are two distinc I points with tji ffe 1/2 and zi 7^2. then I y-i - yi *2 - zi ) 542 Coordinates. In Space Chapter 12 Proofs of Theorems 12,11, 12.12, 12.13 arts assigned as exercises. THEOREM 12 .1 1 If P(x lt y h ft) and Q(xi, y lt * 3 ) are two distinct points, then W = ((*. y.z}:x = xi and y = y t }. THEOREM 12. 12 If P(x lt y x > zi) and Q{x x , y 2t Si) are two distinct points, then ¥@ = {{*, y t z):x = xi and *= %). THEOREM 12.13 If flfa, ^i> *i) and 9(* 2 , if! . a,) are two distinct points, then PQ = {(*, y, s) : y = yi and s = *i). EXERCISES 12,5 ■ In Exercises 1-5, a tine is given in terms of symmetric equations. In eac-h case, find the missing coordinates of P t Q t R so thai they will be points of the given line. m. ib m x e& m. m ). «{3. in. h ) "£3. [Em). «7. a, >, j!(n, s h ) 1 -5 3 2 J nm m, -t), p( m, m, -3), a® m, 1 1 fc(fc*4>f-f-i}- 12.5 Symmetric Equation* for a Line 543 ■ In Exercises 6-1 0, a line is given. To which of the coordinate axes, if any, is it parallel? 10. fa y, z) : x = 3, y = 4} ■ In Exercises 1 1— IS, a Hue is given in terms of parametric equations. Repre- sent the line using symmetric equations. 11. {(,*, y,z): x = 1 -f 3k, r/ = 2 — 2k, z = 3 — 4Ar„ fc is real} 12. {{x, y,z):x = - 1 - k, y = 2 + 4k, z = 5 - k, k is real} 13. ((x. {/, z) : .t = A\ y s 2k t * = 3fc, fc is real} 14. {(x, ^ z) : x = 3 - 4k y y = 2fc, 2 = 1 - k, k is real) 15. {x, y. z) : x = 2 + ft(5 - 2), y = 3 + Jk{5 - 3), 2 9 2 + fc(l - 2), k is real} ■ In Exercises 16-20, a plane and a line are given. Find their point of intersection. lfi. ((x, i Jt z) : 3x + 4y + 55 = 80}, {(*, I/, jj : -|= -| « -|) 17. {{x, j/, 5) : 2x -1- y - 10 = 0} T {(x T r/, s) ; * = 5 and 3 = 3} 18. {{x, ,j,z):x-y + z = 0}, [{x, y, z) ; ^1 = ±^A = i^.} 16. {(x, y, z) : 2x - y + z + 5 = 0}, 20. {{x, j/,s) : x + «/ + 3 + 1 = 0), {{x, $f. 3) : % = y = z) 21. Prove Theorem 12.9, 22. Prove Theorem 12,10. 23. Prove Theorem 12.11. 24. Prove Theorem 12.12. 25. Prove Theorem 12.13. 544 Coordinates in Space Chapter 12 CHAPTER SUMMARY We began this chapter by defining an xj/jb-COORDINATE SYSTEM. Stalling with a pair of perpendicular lines called the x- and the y-axcs, we know that there is one and only one line through their intersection that is perpendicular to both of them* We call this line the z-AXIS, The one and only point that lies on all three of the coordinate axes is called the ORIGIN'. There are three coordinate axes and three coordinate planes. Each axis lies in two of the coordinate planes and is perpendicular to the oilier one. All lines perpendicular to a coordinate plane are parallel to one of the coor- dinate axes. AH lines parallel to a coordinate axis are perpendicular to one and the same coordinate plane. If a plane is parallel to a coordinate plane, then it is perpendicular to a coordinate axis and to all lines parallel to that axis. In our work with coordinates in space we stated and proved the DISTANCE FORMULA THEOREM and slated a theorem regarding parametric equations for a line, These theorems are natural extensions of some theorems for the two-dimensional case in Chapter 11, The Distance Formula and parametric equations for a line are useful tools in solving prob- lems involving solid geometry. Linear equations may be used to represent planes, and combinations of them in what we call symmetric form may be used to represent lines. REVIEW EXERCISES In Exercises 1-8, eight points are given as follows: A(0, 0, 0), 13(3, 0, 0), C{3, 4, 0), D(Q, 4, 0), £(0, 0, 2), F(3, 0, 2), G(3, 4, 2), H(0, 4, 2). 1. Draw a graph of the rectangular solid with vertices A. B, C, D, E. F, G,H. 2. Prove that lABF^ L DBF — / CBF =z LEFC. 3. Find the measure of dihedral angle II-EA-F, 4. Prove that AG = EC ss W =s ¥D. 5. Find the distance from point A to plane CDH. & Find the distance from point A to plane FOIL 7. Find the distance hetwecQ plane ABC and plane EFC2. 8. l^et I. J, K be the midpoints of EE, ItfZ, DC ( respectively. Find the dis- tance between the planes JJK and ECC. In Exercises 9-13, find the distance between the two given points, 9. (Xi, t flt %) and (jfifc t/ 2 , z*). 12. (7, 8, 0) and (10, 12 t 0). 10, (a, b, c) and (d, e, /). 13. (3, 4, 5} and (8, 9, 10). 11. (p, q, r) and (p t q, t). Review Extrclsts 545 In Exercises 14-22, there are two given points: A{1, 2, 3) and #(4. 5, 6)* 14. Express AB using parametric equations and set-builder notation. 15. Express AB using parametric equations and set-builder notation. 16. Express BA using parametric equations and set-builder notation, 17. Express opp AH using parametric equations and set-builder notation. 18. Express opp BA using parametric equations and set-builder notation. 10. Find C on AB if AC = 10 - AB. 20. Kind C on opp AS it AC = 10 * AB. 21. Find the bisection points of AE. 22. Find the two points winch divide AB internally and externally in the ratio ,. 23. Given A(5, 1.1). fl{3, 1, 0), C<4. 3, -2). JJ(6 t 3 T -1}, prove that quad- rilateral ABCD is a parallelogram. 24. Given A(2, 4, 1), B(l, 2, -2), C(5> 0, -2), pro%-e that AABCis a right triangle, 25. Find three noncollincar points on the plane given by the equation 3x + Ay - z + 15 = 0. 26. Find three distinct points on the line given by the symmetric equation x—2 _ y-4 _ z+3 3 ~ 2 5 ' 27. Write an equation of the plane that is perpendicular at (2, 4, — 3) to the line of Exercise 26. 28. Find the x-, y-, and s-intercepts of the plane a where a = {(jr, tj t s) : 3x - 7y - fa + 5 = 0}. 29. Express the line / where I = {(*, sj, z) : x = 1 + 2*, y = 1 - 3fc, z = k,k is reu]}, using symmetric equations but no parameter. 3(1. Consider the line I and the plane or perpendicular to / at C>(0, 0, 0), given as follows: Find two distinct points P and Q on J such that « is the perpendicular bisecting plane of T$, / \ 4 Chapter Erick ttartmoimf Magnum Photot Circles and Spheres 13.1 INTRODUCTION An the title suggests, this chapter is concerned with properties of circles and spheres, some of which you may already have studied in your earlier work in mathematics. The first part of the chapter deals with the intersection proj)erties of a circle and a line in the plane of the circle and the intersection properties of a sphere and a plane. The second part of the chapter is about the degree measure of arcs of a circle and properties of certain angles in relation to arcs, secants, tangents, and chords. We also consider properties of lengths of secant- segments, tangent-segments, and chords. 13.2 CIRCLES AND SPHERES: BASIC DEFINITIONS Up to now, in our formal geometry, we have not discussed "curved figures/' that is, figures not made up of segments, rays, or lines* Perhaps the simplest of these figures are the circle, the sphere, and portions of a circle, called arc*;. We begin with some formal definitions. 548 Circles and Spheres Chapter 1 3 Definition 13.1 (See Figure 13-1.) Let r be a positive num- ber and let O be a point in a given plane. The set of all points Fin the given plane such that OP — r is called a circle. The given point O is called the center of the circle, the given number r is called the radius of the circle, and the number 2r is called the diameter of the circle. A circle is a curve and encloses a portion of a plane. As an illustra- tion, consider a circular disk. The edge of the disk is what we have in mind when we think of a circle. The edge of the disk together with its interior points is what we have in mind when we think of a circular region. Wc will have more to say about a circular region (that is, a circle and its interior) in Chapter 14, Question: Is the center of a circle a point of the circle? Explain, Circle with center O and radius OF = r ^V Sphere with center and radius, OP = r Figure 13-1 Definition 13.2 (See Fij^ire 13-1.) Let r be a positive Dum- ber and lei () be a point in space. The set of all points P in space such that OP = r is called a sphere. The given point O is called the center of the sphere, the given number r is called the radius of the sphere, and the number 2r is called the diameter of the sphere. A sphere is a surface and encloses a portion of space. As an illus- tration, consider a ball. The surface of the ball is what we have in mind when w-e think of a sphere. The surface of the ball together with its interior points is what we have in mind when we think of a spherical region. We will have more to say atout a spherical region (that is, a sphere and its interior) in Chapter 15, 13.2 Circles and Spheres: Banc Definitions 549 Definition 13.3 Two or more coplanar circles, or two or more spheres, with the same center are .said to be concentric. In Figure 13-2, Q is the common center of three concentric circles with radii (plural of radius) r-i, r 2l and r 3 „ respectively. Figure 13-2 Let O be a point in a given plane a. Then we can choose an xy-coor- dinatc system in a such that the origin is at O. Let Cbe a circle in a with center at O and radius r. Let P(x, y) be any point on C as shown in Figure 13-3. Pto, y) Simon 13-3 Then, by the Distance Formula, OP = r = >/(x - 0)- + (y - G) 2 . Therefore r = \/^~+ y 2 and x 2 + y- = r 2 . Converse ly, if F(x, y) is any point such that x 2 + y* = f*, then y/( x - 0) 2 + (y - 0) 2 = r T OF = r, and F is a point of the circle C. Thus, in a given xt/-plane, the circle C with center at the origin and with radius r is given by C = {{x, y) ; x 2 + y 2 = r 2 }. We have proved the following theorem. 550 Circles and Spheres. Chapter 13 THEOREM 13.1 Let an xy-phme be given and let O be the origin and let r be a positive number. Let C be the circle in the sy-plane with center O and radius f. Then C = {(*, y) : ** + r/2 = r*}. Definition 13.4 A chord of a circle or a sphere is a segment whose endpoints are points of the circle or sphere. A secant of a circle or sphere is a line containing a chord of the circle or sphere- A diameter of a circle or sphere is a chord contain- ing the center of the circle or sphere. A radius of a circle or sphere is a segment with one endpoint at the center and the other endpoint on the circle or sphere. In Figure 13-4, C is a circle with center P, and S is a sphere with cent er P. For the circle and the sphere, AB is a chord, 46 is a secant, ED is a diameter (and also a chord), and PQ is a radius. The endpoint of a radius that is on a circle or a sphere (such as point Q in Figure 13-4) is often referred to as the outer end of that radius. Is ED in Figure 13-4 a secant? It follows from Definition 13,4 that a secant is a line that in- tersects a circle (or sphere) in two distinct points. *<* Figure 13-* Note that, in connection with a circle or a sphere, the word "radius" is used in two different ways and the word "diameter" is used in two different ways: (1) each word 1$ used to mean a certain segment and (2) each word is used to mean the positive number that is the length of a segment. Hus should not be confusing because the context in which the word is used should make it easy for you to decide which meaning Ls intended. For example, if we speak of a radius or a diameter of a circle or of a sphere, we mean a segment. If we speak of the radius 13.2 Circles and Spheres: Basic Definitions 551 or the diameter, we in can the number that is the length of a segment. Thus a circle has infinitely many different radii if radius is interpreted to mean a segment; it has just one radius if radius means the length of a segment. Just as we speak of congruent angles, or congruent segments, or congruent triangles, we often speak about congruent circles or con- gruent spheres. How would you define congruent circles or congruent spheres? Does your definition agree with the following one? Definition 13,$ Two circles (distinct or not) are congruent if their radii arc equal. Two spheres (distinct or not) are con- gruent if their radii are equal. Using Definition 13.5, it is not hard to prove that congruence for circles (or spheres) is an equivalence relation; that is, it is reflexive, sym- metric, and transitive. Thus if A, B, C are any three circles (or spheres), it is true that I A=A. 2. IfA^B, thcnBsA. 3. Tf A == B and B s C, then AsC. Reflexive Property Symmetric Property Transitive Property EXERCISES 13.2 In Exercises 1 10, refer to the circle with center F shown in Figure 13-5. Assume that all the points named in the figure are where they appear to l>e in the plane of the circle. Copy and replace the question marks with words or symbols that best name or describe the indicated parts. 1. TJE is called a [7] of the circle. 2. FE is called a [T] of the circle. 3. AH is called a [Tj of the circle. AB could also be culled a (T] of the circle. 4* FG is called a (7] of the circle. S. FG is called a (7] of the circle. & DE is called a [7] of the circle. 7. C is the outer end of the radius [TJ* 8. A is the |7J of the radius |7]. 9. The points named in the figure thai arc points of the circle (that is, on the circle) are |7], 10. The points named in the figure that arc not points of the circle are [?]■ l-i^m. U-.- 552 Circles and Spheres Chapter 13 Li Exercises 11-15, refer to the Sphere with center Q shown in Figure 13-6. Copy and replace the question murks with words or symbols that best name or describe the indicated parts. Figure 13-9 11. QR is a [Tj of the sphere. 12. If S, Q, T are collincar, then ST is a [TJ of the sphere, 13. KS is a \?} of the sphere, 14. RT h a [?] of the sphere. 15. Points T| are outer ends of given radii. 16. Prove that if two circles are congruent, then a diameter of one is con- gruent to a diameter of the other. 17. Prove that congruence of circles is an equivalence relation, (You will need to use Definition 13.5 and tbe theorem that congruence of seg- ments (radii) is an equivalence relation.) In Exercises 18-25, refer to the circle with center at the origin of Lhc xy- conrdinatc system shown in Figure 13-7, P{x t o) is a point on the circle. Figure 13*7 ! < ^y j* P( "K / - V * { \* q -6 1-4 .3.3-1 t 12 3 4 1 6 ' \ -- 1 / 1 1 13.2 Circles and Spheres: Basic Definitions 553 18. Use the Distance Formula and express the distance between O and P in terms of x and y. 19. Write an equation like that in Theorem 13. 1 of the circle with center at the origin and radius 5. 20. Write the coordinates of the points on the x- and y-axes that arc on the circle of Exercise 19. Check your answers by substituting the coordi- nates in the equation of the circle. 2L Is A = (3, i) u point of the circle of Exercise 19? 22. Is B = ( — 4, 3) a point of the circle of Exercise 19? 23. Is K = (l t 2%/^) a point of the circle of Exercise 19? 24. Is R = (—2, 4.5) a point of the circle of Exercise 19? 25. Write the coordinates of two more points (different from those in Ex- erases 20-24) that lie on the circle of Exercise 19. 26. Given the ,t [/-coordinate system shown in the figure with A = (1,2), P b (x, y) y and AP — 5, use the Distance Formula and express the dis- tance between A and Fin terms of x and y, ' '>' , . 4 «*.*j 4 *\ 4(1.2) x * -2-1 12 3 4 s a ■ i 27. In Exercise 26, write an equation like that in Theorem 13.1 of the circle with ecntcr at A and radius AP = 5. Is the point S = (4, 6) a point of this circle? Write the coordinates of at least three more points that are on this circle. 25. Which of the following are points of C as {{%, y) : x 2 + y 2 = 100}? (a) (a 10) (b) (-6,8) (c) (4,9) (d) (2^ -2\/I5) 29. Whet is the nidi us of the circle of Exercise 28? 30. Find five more points of C in Exercise 28. Exercises 31-34 concern the set C, where C - {(x, y) : i* + y 2 = 16). 31. Is C a circle? Why? 32. Find x if {x, 3) is a point of C. (There are two possible values for a ; 33. Find y if (4. y) is a point of C. 34. Can you find x so that (x, 5) is a point of C? Explain. 554 Circles and Spheres Chapter 13 ■ Exercises 35-37 concern the circle C» where C = {(x, y) : x 2 + y z = 36). 35. What restrictions on x and t/ would give only the part of the circle in quadrant 1? 36. What part of the circle would you be considering under the restriction x 2 + y 2 = 36 and x < 0? 37. What restrictions ou x and y would give the intersection of C and quad- rant 111? 38. challenge problem. Let an xi/-plane with A = {h, k), F = (x t y), AP = r > be given. Write an equation in terms of x, y, h f k t r for the circle with center al A and radius AP = r. 13,3 TANGENT LINES If you look at a drawing of a circle in a given plane, it is easy to see that the circle separates the points of the plane not on the circle into two sets. One of these sets consists of those points of the given plane that are "inside" the circle and the other set consists of those points of the given plane that are "outside" the circle. Definition 13.6 (See Figure 13-8.) Let a circle with center O and radius r in plane a he given. The interior of the circle is the set of all points P in plane a such that OP < r. The exterior of the circle is the set of all points P in plane « such that OP > r. Home 13-S It is clear from Definitions 13,1 and 13.6 that if F is anv point in the plane of a circle with center O and radius r, then P is on the circle {OP = r), or P is in the interior of the circle (OF <[ r), or P is in the ex- 13.3 iMpii Lines 555 terior of the circle {OP > r). Wc sometimes say "Pis inside the circle' or "P is outside the circle*' when we mean "P is in the interior of the cirde" or "Pis in the exterior of the circle. 1 ' respectively. Figure J 3-9 shows an ^-coordinate system whose origin is the center of a given circle C with radius r. The figure also includes expres- sions for C, its interior and its exterior, in terms of coordinates using set- Wilder notation. Inferior of C =• {(*; Jf) F.xlerk* at C = {(*»jr) «» + 9* = »*) •* + V* > **) Figure 13-9 Definition 13.7 If a line in the plane of a circle intersects the circle in exactly one point, the line is called a tangent to the circle and the point is called the point of tangeney,or die point of contact. Wc say that the line and the circle arc tangent at this point. J f a segment or a ray intersects a circle and if the line that contains that segment or ray is tangent to the circle, then the segment or ray is said to lx* tangent to the circle. In Figure 13-10, if Ms a line in the plane of circle C and if the in- tersection of I and C Is just the one point P, then I is tangent to C at P. Figure 13-10 5 56 Circles and Spheres Chapter 13 Given a circle and a line in the plane of the circle, what are the possi- bilities with regard to the intersection of the line and the circle? Figure 13-1 1 suggests that there arc only these three possibilities: the inter- section of the line and the circle may be the empty set, or it may be a set consisting of exactly one point, or it may be a set consisting of exactly two points. cnl=\P\ t blgwo 13-11 That these arc the only three possibilities can be verified in the fol- lowing way. Suppose that wc are given a circle C with center in plane a and that P is a point in plane a. Then P is outside the circle, as shown in Figure 13-1 la, or P is on the circle, as shown in (b) t or P is inside the circle as shown in (e). If Pis outside the circle, we shall show that the unique line I in plane cr such that OP _^ I at P does not intersect the circle. If P is on the cir- cle, wc shall show that the unique line / in plane a such that OP ± I at F intersects the circle in exactly one point (hence / is tangent lo C at P). If f is inside the circle, then either P = O or P=£CK1( P / 0, we will show that the unique line I in plane a such that OP I I at P inter- sects die circle in exactly two points which are equidistant from P, Finally, if P = O, we will show that any line I which contains P inter- sects the circle in exactly two points. We are now ready for the follow- ing theorem. 33.3 Tangent Lines 557 THEOREM 13.2 Given a line I and a circle C in the same plane, let O be the center of the circle and let P l>e the foot of the per- pendicular from O to line I, L Every point of 2 is outside C if and only if P is outside C 2. I is tangent to C if and only if P is on C 3. I is a secant of C if and only if P is inside C. Proof: Let r be the radius of C and let OP = a. We select an ^-coor- dinate system in the plane of C and I with the origin at O s with the y-axis parallel to I and with P on the nomiegative ar-asis. Then and P=(a,0)> Z= {(*,*/) :* = a}. Proo/ 0/ i; Suppose we are given that ? is outside C as shown in Fig- ure 13*12; then a > r]> 0. Why? It follows that a 2 > r 2 and hence Q 2 + y 2 > **• Therefore all points (fl* y) are outside C, Since t-{{x i y):x=a} = {(a, y) : y is real}, it follows that all points of J ate outside C. figure 13-12 Now suppose that I is outside C; then every point of I, including P % is outside C and the proof of (1) is complete. 558 Circles and Sphert* ::>spter ".3 Proof of 2: Suppose wc are given I hat F is on C as shown in Figure 13-13; then a = r. Why? The intersection of C and I is Ifa y) : x* -f i/ = r*} fl {(*, y):x = u} = {(a. y):y* = Q}> c Ij Pte.D) FlRUIT 13-13 Since zero is the only number whose square is zero, it follows that y = 0. Therefore the only point of intersection of I and C is P{m 0). Thereforc I is tangent to C at P. Why? This proves that I is tangent to C if P is on C, Now suppose it is given that I is tangent to C. Then J and C have exactly one point in common. This means that I H Cis a set consisting of exactly one point. But C = {{*, y) : x* + f = i*}, and IDC= {{x, \j) : x* 4- J/ 2 = r 2 and* = a} -{fey):* 4 E/ 2 = r2}. If (a, tj) with t/ ^ Q is a point of Z fl C, then (a, -y) is also a point of I H C and there are two distinct points in I D C. Since there is only one point in / fl C, it follows that y = and a z = #, Since a > 3 it follows that a m f and the one and only point in I H C is the point HA 0)* This proves that / is tangent to C only if P is on C, 13.3 Tangent Un« 559 Proof of 3: Suppose w© are given that P is inside C as shown in Figure 13-14a or b. Recall that O is the center of the given circle C Then P = («, 0) where < a < r> an< I n G = {(*, y) : x = a and * 2 + t/ 2 = r 2 } Since < a < r, il follows that r* - o 2 > 0, y/fi-d* ^ - Vr^fl* and the points (a, "v 71,2 — ° 2 ) an ° 1 (*>■ — V 4 — ° 2 ) are distinct points. This proves that / is a secant of C if P is inside C. Suppose now that I is a secant of C; then / intersects C in two dis- tinct points which we have just shown to be (a. yV 2 — a 2 ) and {a> — y/f& — « a ). This means that r 2 > « a . (Why is it that we cannot have r a = a 2 or r 2 < a 2 ?} Since r > and « > 0, it follows that r > a and hence that OP <C ft Therefore F(a, 0) is inside C and the proof is complete. Now that we have proved Theorem 13.2 we proceed to state our first basic theorems on tangents and chords. To prove some of these theorems wc need refer only to Theorem 13.2 and see which of parts (1), (2), or (3) apply. 560 Circles and Spheres Chapter 13 THEOREM 13.3 Given a circle and a line in the same plane, if tlie line is tangent to the circle, then it is perpendicular to the radius whose outer end is the point of tangency. Proof: Let C be the given circle with center O, let line I be tangent to C at R, and let P be the foot of the perpendicular from O to /. it fol* lows from part 2 of Theorem 13.2 that P is on C. (See Figure 13-15.) The figure shows R and P to be distinct points. We shall prove that they are the same point. If Pi=R, then OR is the hypotenuse of right triangle AOPR and OR > OP, But R and P are both on C. Therefore OR = OP. Since this is a con- tradiction, it follows that P = R. Since I ±OP at P t it follows that I ± OR at R, and the proof is complete. Figure 13-15 Our next theorem is the converse of Theorem 13.3. THEOREM 13.4 Given a circle and a line in the same plane, if the line is perpendicular to a radius at its outer end, then the line is a tangent to the circle. Proof: Assigned as an exercise. THEOREM 13.5 A diameter of a circle bisects a chord of the circle other than a diameter if and only if it is perpendicular to the chord. Proof: Let a circle C with radius r, center O, diameter AJ3, and chord RQ be given. Wc must prove two things. Ll£AB±KQ t then A75 bisects ftg. 2. If AB bisects RQ, then A~B ± RQ. Proof of 1 : Choose an ^-coordinate system in the plane of the given circlc_C such that the origin is at the center 0,A = { - r, 0), B = (n 0), and RQ is perpendicular to KB at P(a, 0), where < a < r as shown in Figure 13-16, Then AB is a diameter of the circle. "Why? 13.3 Tangent Lirm 561 Figure 13-16 Suppose /? and {) are named so that R is in the first quadrant. Then (see the proof of part 3 of Theorem 13,2) R = (a, y/& — a 2 ) Q = (a, — ^/r* — a 1 }. Therefore and FR = JVf«-fl 2 -0| = V* " o* Pp = p - (- Vf2 - a*)j = v^^ 1 ^- Therefore Fit = FQ and AB bisects ff^ at P. This proves that if a di- ameter is perpendicular to a chord, then it bisects the chord. Proof of 2; Choose an ary-coordinate system in the plane of the given circle C such that the origin O is the center of the circle, A = ( — r, 0), B = (r, 0), and the chord EQ (not a diameter) intersects AB at P{a, 0) with < a < r, (See Figure 13-17.) Figure 13-17 562 Circles and Spheres Chapter 13 Suppose that P is the midpoint of RQ. Then If R = (x^ ^) and Q — (** Sfe)* *■ nave Pfl = Pp, (*! - a)> + kl! = (* 2 - a ) 2 + y 2 2 , (Why?) *i 2 - 2*ia + 0* + y,a = Xi 2 - 2x& + a 2 + y#. But xi 2 + tji 2 = f 2 - *a a + yz 2 - (Why?) Then — 2x\a a — 2^ 2 a> *i = **. RQ is a vertical line, and RQ I AB, This proves that if a diameter bisects a chord that is not a diameter, then the diameter and the chord are perpendicular. This completes the proof of Theorem 13.5, THEOREM 13,6 In the plane of a circle, the perpendicular bi- sector of a chord contains the center of the circle. Proof: In the proof of part 1 of Theorem 13.5, AB is the perpendic- ular bisector of chord BQ in the plane of the given circle by the defi- nition of the perpendicular bisector of a segment in a plane. Since An contains O, the center of the given circle. Theorem 13.6 is proved* TIJEOREM 13.7 Let a circle C and a line I in the plane of the circle be given. If I intersects the interior of C, then I intersects C in exactly two distinct points. Proof-. Theorem 13.7 follows from part 3 of Theorem 13,2. The de- tails of the proof are assigned as an exercise. THEOREM 13,8 Chords of congruent circles are congruent if and only if they are equidistant from the centers of the circles. Proof: Let C" and C be the given congruent circles with centers P and F and radii f and f, respectively. Then r = f. Why? Let AB and A'B' be chords of the given circles C and C, respectively. Suppose, first that the distance from Ali to the center of circle C is zero, that is. AB is a diameter of C, and suppose that AU ^= A'B'. Then AB = A'B' = 2r = If and ~ATE is a diameter of circle C Therefore AB and A'B' arc equi- distant from P andF (th e dista nces being in this case). Conversely., if the distances of AB and A'B' from P and F are 0, then AB and /PF are diameters of C and C. It follows that AB = 2r = 2/ = A'B 1 ; hence AB === A'B', 13.3 Tangent Lines Now suppose that AB is not a diameter of C. If AB s A'B\ it fol- lows that A'B' is not a diameter of C, Whv? Let F be the foot of the perpendicular from P to AB and let F be the foot of the perpendicular from F to A'B' as shown in Figure 13-18. Then by Theorem 13 .5, A** = l 2 AB and A'F = JA'B'. 5*3 Figure 1 3- IS By hypothesis, "Therefore We have AB = A'B'. AF = A'F. AP = A'F. (Why?) Since AAPF and AA'FF are right triangles, it follows that AAPF ss AA'FF by the Hypotenuse-Leg Theorem. Therefore PF = FF and hence AB and A'fr are equidistant from F and F, Conversely, if KB and A'B' are equidistant from P and F, that is, if Pf?=FF#0 ( ft follows that AAPFss AA'FF by die Hypotenuse-Leg Theorem, (Show that AAPF & AA'FF if PF = FF.) Therefore AF = A'F. But by Theorem 13.5, AF = and A'F = ^A'F. AB = A'B' and ^EssA'/?'. This completes the proof of Theorem 13.8, It should be noted that the congruent circles of Theorem 13.8 could be the same circle, in which case the theorem still holds. 564 arete* and Spheres Chapter 13 Figure 13-19 shows two different examples of two circles tangent to the same line at the same point In Figure 13-19& the centers A and A' of the two circles are on the same side of the tangent line I and the circles are said to be internally Umgent, In Figure 1 3- 1 9b the centers B and B' of the two circles are on opposite sides of the tangent line n and the circles are said to be externally tangent. [-Iml-.-ujIIj, Rutgeal - m Figure 13-19 Our formal definition follows. Definition 13.8 l\vo circles are tangent if and only if they are coplanar and tangent to the same line at the same point If the centers of the tangent circles are on the same side of the tangent line, the circles arc said to be internally tangent. If their centers arc on opposite sides of the tangent line, the circles are said to be externally tangent. EXERCISES 13.3 Exercises 1-10 refer to the circle C •= {(*, tf) : x 3 ■§■ tf = 64}. In each ex- ercise, the coordinates of a point arc given. Tell whether the point is en the circle, in the interior of the circle, or in the exterior of the circle. h (0, -8) 4. (-4,7) 7. (0, 8) 10. (6. -2\/7) 2, (3, 5) 5. (-4v^,4) 8. (8, -1) 3. (-7,3) G, (4V3, -4) 9. (-6, -6) 11. Find the endpoints of two distinct diameters of the circle C = {(*» y) : ** + y* = ea } . 12, Use set-builder notation to express the set of points in the exterior £ of the circle C = {(*, tj) : x-' + tf = 64}. 13,3 Tangent Un*s 565 13. Use set -builder notation to express the set of points in the interior I of the circle C ■ {(x, y) : x 2 + y 2 = 64}. 14. Prove that the center of a circle is in the interior of the circle. Exercises 15-20 refer to the circle C" with center tit (0, 0) and passing through the point ?(- 5, 12). 15. Find the radius of the circle. 16. Use set-builder notation to express the set of points on the circle. 17* Find eight distinct points that are on the circle 18. Find two distinct points that are in the exterior of the circle. 19. Find two distinct points that are in the interior of the circle. 20. If / is the tangent line to C at P(— 5, 12). find an equation for L (Hint; I — UP at P, Find the slope of I and use the Point- Slope Form of an equation,) 21. Ut the circle C = {(*, y) : x 2 -f y* = 36} and the line I = {(x, y) : x = 3} be given, (a) Does I intersect C? (b) If / intersects C, is I a tangent line or a secant line? (c) If I intersects C, find the coordinates of the poiutjs) of intersection. 22. Let the circle C = {(x, y) : x* + y 2 = 25} and the lines Ei = {(*.y) : x = -5}, h= {(*,$f) : y = 2*}, h = {fry) : * = 7) he given. (a) Which of the lines intersect the circle? (b) Which line is tangent to the circle? What are the coordinates of the point of Langency? (c) Which line is a secant? What are the coordinates of the points of intersection of the secant line and the circle? 23. Prove Theorem 13.4. 24. Prove Theorem 13.7, 25. Copy and complete: A tangent to a circle is JT] to the radius drawn to the point of contact. 26. Copy and complete: If a diameter is perpendicular to a chord, then it [?] the chord. 27. Copy and complete: If a diameter bisects ft chord other than a diameter, then it Is |3. 28. Copy and complete: In the plane of a circle, the perpendicular bisector of a chord contains the [T]. 29. In a circle with r ad his 13 in., how long is a chord 5 in. from the center of the circle? 30. In a circle with diameter 12 cm., how long is a chord 4 cm, from the center of the circle? 3L Find the radius of a circle if a chord 8 in. long is 3 in. from the center of the circle. 566 Circles and Spheres Chapter 13 32, How far from the center of a circle with a radius equal to 25 is a chord whose length is 30? 33. In the figure. FQ is parallel to I which is tangent to the circle at F. The center of the circle is R and FQ bisects RF at Jtf. If FQ = IS, find BP, [Hint: Let RF = 2x. Then RP = 2x and RM = x.) w Jl 34. If AH is a diameter of a circle and if lines t t and i-* arc tangent to the circle at A and B t respectively, prove h \\ £&. 35. Prove that the line containing the centers of two tangent circles con- tains the point of tangeucy. (See Figure 13-19.) 36. The figure below at left shows two concentric circles, AB is a chord of die larger circle and is tangent to the smaller circle at AC Prove that M is the midpoint of A~B t 37. In the figure above at right, T is a point in the exterior of the circle with center P. T\vo distinct tangents are drawn from T to the circle with points of contact A and B. Prove that TF bisects LATB and that AT = BT. 3S. In Exercise 37. if the radius of the circle is 9 and FT = 15, find AT. 39. challenge problem. In Exercise 37 if the radius of the circle is 9 and FT = 15, find AH. 13.4 Tangent Plana* 567 40. CHALLENGE PROBLEM. Show thilt the lillC I = {(x. If) I 3x 4- 4(/ = 25} i8 tangent to Uic circle C = {(*, tf) : ** + y 2 = 25) and find the co- ordinates of the point of tangency. 41. challenge problem. Prove that no circle contains three col linear points. {Hint: Use Theorem 13.6.) 13.4 TANGENT PLANES In Section 13.3 we studied relations between lines and circles in a plane. In this section we study relations between planes and spheres in space. There is a close analogy between the definitions and between the theorems of the two sections. Onr first theorem of this section is analogous to Theorem 13.1, In Chapter 12 you learned how to find the distance between two points in space by introducing a three-dimensional coordinate system (called an $j/£-coordinatc system). We shall use this Distance Formula to develop an equation for a sphere in an xy£-coordtaate system. Let O be a point and r a positive numl ier. Let S be the sphere with center O and radius r. Suppose an ^-coordinate system has been set up with O as the origin. (See Figure 13-20.) Then Pfc tj, %) is a point of S if and only if OF = s/{x - 0)2 + (y - 0)8 + (Z - 0)-' = r s V* 2 + tf 3 + « a = r, * a + y 2 4- z 2 = r 2 . P(x,y,»ji Figure 13-30 We have proved the following theorem. 568 Circles and Spheres Chapter 13 THEOREM 13.9 Let be a point, r a positive number, and S the sphere with center O and radius r. Given an jcy.s-coordmate system with origin O, S = {(ar, y, z) ; x 2 + y 2 + s 2 = r 2 }. Definition 13M (See Figure 13-21.) Let a sphere with center and radius r be given. The interior of the sphere is the set of all points P in space such that OF << r. The exterior of the sphere is the set of all points P in space such that OP>r. e J:.-,'Tk,r *\. Exterior <> -^ Figure 13-21 In view of Definition 13-9 and Theorem 13.9. if S is a sphere with radius r and center at the origin of an 3fys-coordinate svstem, then S = {(*, if, z) : X 2 + tf + *2 = r 2 } Z= {(*,y f *):a* + if + ^<ia} and where / is the interior of the sphere and £ is its exterior. Before reading Definition 13.10, try to form your own definition of a plane tangent to a sphere. See Figure 13-22. Figure 13-23 13.4 Tangent Plants 569 Definition 13.10 If a plane intersects a sphere in exactly one point, the plane is called a tangent plane to the sphere. The point is called the point of tangent- \\ or the point of contact, and we say that the plane and the sphere are tangent at this point. You have seen that there are three possibilities with regard to the intersection of a line and a circle in the same plane. Figure 13-23 suggests three possibilities with regard to the intersection of a plane and a sphere. For each part of Figure 13-23. S is a sphere with center O and P is the foot of the perpendicular from O to plane a. Figure 1 3-23a suggests that if P is in the exterior of the sphere, then all of plane a is in die exterior of the sphere and S n n = . Figure 13-23b suggests that if jP is on the sphere, then a is tangent to the sphere at P and Sfla = {P}. Figure l3-23c suggests that if P is in the interior of the sphere, then the intersection of the sphere and plane ot is a circle C with center P, that is, S n a = C. a (a) (b) (O Figure 13-33 The following theorem concerning the intersection of a plane and a sphere is analogous to Theorem 13.2 concerning the intersection of a line and a circle in a plane. It can be proved using a three-din ie us it mal coordinate system in much the same way that Theorem 13.2 was proved using a two-dimensional coordinate system. You will be asked to write a proof of Theorem 13.10 in the Exercises. THEOREM 13.10 Given a sphere S with center O and a plane a which docs not contain O, let P be die foot of the perpendicular from O to a. L Every point of a is in the exterior of S if and only if P is in the exterior of S. 2* a is tangent to S if and only if P is on $♦ 3. a intersects S in a circle with center P if and only if P is in the interior of S, 570 Circles and Spheral Chapter 13 1HEOREM 13. J J Let a sphere S with center O and radius r and a plane a be given. If the intersection of S with a contains the center O of the sphere, then the intersection is a circle whose center and radius are the same as those of the sphere. Proof: Let a sphere S with center O and radius r be given as shown in Figure 13-24. Let a be a plane that contains O and intersect*. S. Since S is the set of all points of space whose distance from O is r, the inter- section of S and a is the set of all points P of a such that OP = r. By definition, this set is the circle in plane a with center O and radius r. Thus the circle has the same center and radius as the sphere, and the proof is complete* Figure 13-24 Definition 13.11 A circle that is the intersection of a sphere with a plane through the center of the sphere is called a great circle of tile sphere. The next two theorems can be easily proved using Theorem 13, 1 1. THEOREM 13.12 The perpendicular from the center of a sphere to a chord of the sphere bisects the chord. Proof: The endpoints of the given chord and the center O of the given sphere determine a plane O. The intersection of plane a with the sphere is a great circle: with center O and having the same chord as the given chord- It follows from Theorem 13.5 that the perpendicula: Irom ( Mo the given chord bisects the chord. THEOREM 13.13 The segment joining the center of a sphere to the midpoint of a chord of the sphere is perpendicular to the chord. Proof: Assigned as an exercise. 13.4 Tangent Planes 571 Our last theorem of this section is analogous to Theorems 13.3 and 13.4. THEOREM 13. 14 A plane is tangent to a sphere if and only if it is perpendicular to a radius of the sphere at its outer end. Proof: Let ft sphere S with center O and radius OF be given. Let « tie the given plane. There are two parts to the proof. 1. If UP A a at P, then a is tangent to S at P. 2. If a Is tangent to 5 at P y then OP _ a. Proof of h We are given that OP J- a al P as shown in Figure 13-25. Let R be any point of a different from P; then OP X M (Why?) and AOPR is a right triangle with the right angle at P. Therefore OP < OR. Why? Therefore ii is a point in the exterior of S. Why? It follows that P is the only point of a that belongs to boil a and S; hence 0! is tangent to S at P. Figure 13-23 Proof of 2: We are given that a is tangent to S at P. We shall use an indirect proof to show that OP ± a, Suppose, contrary to what we want to prove, that OP is not perpendicular to a. Let Q be the foot of the unique perpendicular from O to a. Then OQ < OF. Why? There- fore Q is in the interior of S, Why? It follows from part 3 of Theorem 13.10 that a intersects S in a circle. But this contradicts the hypothesis that a is tangent to S; that is, the intersection of a and S is exactly one point. Therefore our supposition that UP is not perpendicular to a is incorrect and we conclude that OP _L a. This completes the proof of Theorem 13.14, 572 Circles and Spheres Chapter 13 EXERCISES 13.4 L I*et a sphere with radius 10 in, lie given, A plane 6 in. from the center of the sphere intersects the sphere in a circle. Find the radius of the circle. 2. Given a sphere S with center O, let a and B be two planes equidistant from O and such that a intersects S in a circle C and /? intersects S in a circle C. Prove that C is congruent to C. {Hint; Let F be the foot of the perpendicular from O to o and let F be the foot of the perpendicular from O to jS. Let G be a point on C and C" be a point on C. Prove that &Omm &OFC.) 3. In Exercise 2, is it necessary for « and /? to be parallel planes? 4 Prove that if the circles of Intersection of two planes with a sphere are congruent, then the planes are equidistant from the center of the sphere. 5, Let a sphere with radius 12 be given, A segment from the center of the sphere to a chord and perpendicular to the chord has length 8, Find the length of the chord, §. A sphere with center C is tangent to plane a at P. AS and CD are lines in plane a which contain P. In what way is CP related to AJS? To CO? Draw a figure which illustrates the given information. 7. Prove Theorem 13.13. {Hint: Give a proof similar to that of Theorem 13,12.) 8. Given a sphere S with center P as shown in the figure. HAS and CD are chords of S which are equidistant from F, prove that AB 5s W and LABP m Z CDP. {Hint: Use Theorem 13.8.) Given that AC and BD are perpendicular diameters of a sphere, prove that ABCD is a square. 13.4 Tangent Planet 573 10. Given that AC and BD are distinct diameters of a sphere, prove that ABCD is a rectangle. 11. Let a sphere S with center P as shown in the figure be given. C is a great circle of S. R is a point on C and T is a point on S, but T is not on C. If ml HPT = 60, prove that ARPT is equilateral. 12. Let a sphere H and a plane a tangent to S at point A be given. Let plane be any plane other than a which contains A. (See the figure,) (a) Prove that plane fi intersects sphere S in a circle C. (b) Prove that piano /? intersects plane a m a line /. (c) Prove that t is tangent to C at A. (Hint: Suppose I intersects C in a second point Q, Then Q is on S (Why?), and hence a intersects S in a second point Q. Contradiction?) 574 Circles and Spheres Chapter 13 ■ Exercises 13-22 refer to the sphere S= [&$$%& + & +&mM}< In each exercise, given the coordinates of a point, tell whether the point is on the sphere, in the interior of the sphere, or in the exterior of the sphere. 13. (0,0,8) 18* (-6, -4,2V3) 14. (0, -8,0) 19. (4,-4, -8) 15. (4 3, 5) 20. (5, 0, -6) 16. (7,2,3) 21. (_2,2 X /B,0) 17. (-4,5,6) 22. (4.3V$ -2) 23. Find the endpoints of three distinct diameters of the sphere $= {{x f if,z):x* + f + J = lQa). ■ Exercises 24-28 refer to the sphere with center at (0, 0. 0) and containing the point (4 t 4* 2). 24. Find the radius of the sphere. 25. Use set-builder notation to express the set of points an the sphere, 26. Find the coordinates of two distinct points that are on the sphere. 27. Find the coordinates of two distinct points that are in the exterior of the sphere. 2& Fiod the coordinates of two distinct points that are in the interior of the sphere. 29. challenge PROBLEM. Theorem 13.14 could be called a restatement of part 2 of Theorem 13.10* Complete the proof of the following restate- ment of part 3 of Theorem 13.10. The intersection of a plane and a sphere is a circle whose center is the foot of the perpendicular from the center of the sphere to the plane if and only if the foot of the perpendicular is in the interior of the sphere. 30. 13.4 Tangent Plane* 575 Proof; 1 ,et a sphere S with center O and radius r and a plane a be given. Let P he the foot of the perpendicular from O to a as shown in the figure. There are two things to be proved, (a) If OP < r, then or H S is a circle G with center P. (1>) If a n S is a circle with center F, then OF < r. Let X be any point of the intersection of a and S. To complete the proof of (a) you need to show that FX is a constant for all points X in the in- tersection of a and 5. To complete the proof of (b) you need lo show thai if FX is a constant for all points X in the intersection of ct and 8, then OP<r. CHALLENCF problem. Complete the proof of the following restate- ment of part 1 of Theorem 13.10. The intersection of a plane and a sphere is the empty set if and only if the foot of the perpendicular from the center of the sphere to die plane is in the exterior of the sphere Proof: Lei a sphere S with center O and radius r and a plane « be given. Let P be the foot of the perpendicular from O to <* as shown in the figure- There are two things to be proved. (a) If OP > r t then all points of « are in the exterior of the sphere. (b) If all points of « are in the exterior of the sphere, then OP > r, 31. challenge problem. See the proof of Theorem 13,2. Prove Theorem 13.10 in a similar way using an xip-coordinate system. 576 Circle and Spheres Chapter 13 13,5 CIRCULAR ARCS. ARC MEASURE Thus far in this chapter we have treated circles and spheres in a similar maimer. In the remainder of this chapter, we limit ourselves to the consideration of topics relating to circles only. The reason for this is that the treatment of the corresponding topics for spheres is too com- plicated to consider in a first course in geometry. We begin with some definitions. Definition 13.12 An angle which Is coplanar with a circle and has its vertex at the center of the circle is called a central angle. Figure 13-S6 In Figure 13-26, P is the center of Ihe given circle and AT) is a di- ameter. LAFB is a central angle. Name three more central angles shown in trie figure. Definition 13.13 (See Figure 13-27.) If A and B are distinct points on a circle with center P and if A and B are not the end- points of a diameter of the circle, then the union of A, B, and all points of the circle in the interior of LAPB is called a minor arc of the circle. The union of A, B, and all points of the circle in the exterior of LAPB is called a major arc oi the circle. If A and B are the endpoiuts of a diameter of the circle, then the union of A, B, and all points of the circle in one of the two halfplanes, with edge AB t lying in the plane of the circle is called a semicircle. 13,5 Circular Arcs, Arc Measure 577 ^i ic."n Figure 10-27 From Definition 13.13, an arc of a circle is either a minor arc, a major arc, or a semicircle. The points A and B in Definition 13.13 are called the endpoints of the arc. Notation. We may denote an arc with endpoints A and B by the sym- bol AjB, which is read "arc AB." However, it should be noted that the symbol AB is ambiguous unless the word "minor" or the word "major" is used in connection with the symbol. For example, in Figure 13-27, we may speak of the minor AjB or the major Aft* Also, semicircle AB is ambiguous since there are two semicircles with endpoints A and B. One way to avoid confusion as to which arc is meant is by choosing an interior point of the arc in question (tliat is, a point of the arc other than its endpoints) and using this third point in naming the arc. Thus, in Figure 13-28, w© may speak of minor AXB or simply AXB. Similarly, we may speak of major A YB or simply A YB, semicircle CXB or semi- circle CYB. Figure 13-2S =78 Circles and Sphtret C-iap:^ 13 It is clear that for each pair of distinct points A, B on a circle there are two arcs which have these points as endpoints. If AB is a minor are, we sometimes say that major AB is tho corresponding major are, or that major AB corresponds to minor AB, or that minor AS corresponds to major AB, In Figure 13-28 name the major arc that corresponds to minor BX. Name the minor arc that corresponds to major ABY. When we speak of the measure of a segment, we mean the nnm- bcr that is the length of the segment. However, this is not true of arcs. That is, when we speak of the measure of an are, we do not mean the "length" of the arc, since length has not been defined for anything except segments. If AXB is a minor arc of a circle with center F, we say that / APB is the associated central angle with respect to AXB. The measure of a minor arc of a circle is related to the degree measure of its associated central angle. We make the following definitions. Definition 13, 14 II AXB is any arc of a circle with center P t men its degree measure (denoted by mAXB) is given as follows: 1. If AXB is a minor arc, then mAXB is the measure of the associated central angle; that Is, mAXB = mZ APR 2. If A XB is a semicircle then tnAXB = ISO. 3. If AXB is a major arc and AY& is the corresponding minor arcs then mAXB = 360 - niAYB, Figure 13-29 shows a circle with center i J . If mlAPB = 50, then and mAXB = 50 mXyS = 360 - 50 - 310. 13.5 Circular Arci, Arc Measure 57*3 Figure 13-19 If BC is a diameter of the circle, what is mCXB in the figure? What is mCYB? Hereafter we shall call mAXB simply the measure of arc AXB with the understanding that wc mean the degree measure of the arc. Is the measure of a minor arc always less than 1 80? Why? Ts the measure of a major arc always greater than 180? Why? Can the measure of an arc be zero? Why? Note that the measure of an arc does not depend on the size of the circle which contains the arc. Figure 13-30 shows 'Concentric circles with center P. with mCYD = mAXB = mLAVB = mlCPD = 35. Figure 13-30 Given three points A, B, C such that B is between A and C, we know that AB 4- BC = AC. Suppose we are given an arc ABC (that is, B is a point, but not an end- point, of AC). It seems reasonable that inAB + JtiBC = rnAFC. We state this as our next theorem. 5 BO Circtas and Spheres Chapter 13 THEOREM 13.15 {Arc Meamre Addition Theorem) If A, B, C are distinct points on a circle, then mXlid = mAB + mBC Proof; Followi ng is a plan for a p roof considering seven possi ble cases as shown in Figure 13-3 1 . In each case, V is the center of the circle and the assertion of the theorem follows from the listed equations. This plans a complete proof since there are no other cases. c A\ U-.-C J Ciae2 Cosu U Cut? Cast I Quti Hgur* 1331 Case 1, ABC is a minor arc. Then AB and BC are minor arcs, mABC = mlAVC mAB = ml AVB mlRm ml BVC m I AVC = m I AVB + ml BVC. Case 2. ABC is a semicircle. Then A B and BC are minor arcs, mABC = 180 mAB = mlAVB mBC- ml BVC 180 = m I AVB -+- m L BVC. 3. ABC is a major arc, and AB and BC are minor arcs. Then mABC = 360 - mlAVC mAB = m I AVB mBC = ml BVC 360 = ml AVB + ml BVC + mlCVA. 13.5 Circular Arcs, Arc Manure a 81 Case 4. ABC is a major arc, AB is a minor arc, and BC is a major arc. Then mABC = 360- mZ.AVC mAB = mLAVB mBC = 260- mLBVC mLBVC = mLBVA + mLAVC. Goa& 5. AJ3C is a major arc, AB is a minor arc, and BC is a semicircle. Then mABC = 360 - mLAVC mAB = mLAVB m£C= 180 180 = mZAVB + mLAVC Case 6. ABC is a major arc, AB is a semicircle, and BC is a minor arc. Then mABC = 360- mLAVC mAB = 180 mBC = mLBVC 180 = mLAVC + mLBVC. Case 7. ABC is a major arc. AB is a major arc, and BC is a minor arc. Then mABC = 360- mLAVC mAB = 360 -mLAVB mBC = mLBVC mLAVB s mLAVC + mLCVB. Note in the situation of Theorem 13, 15 that if D is a point of the circle not on ABC, then mCD/1 = 360 - mABC and mAB + mBC + mGDA = mA5c + mCDA = 360. In other words, if A. B, C are three distinct points on a circle that par- titions the circle into arcs, AB, BC, CA, intersecting only at their end- points, then mAB + mBC + inCA = 582 Circles and Spheres Chapter 13 This idea may be extended to any numl>er of points on a circle, as in the following corollary. COROLLAR Y I.'i, 15,1 U A i, A 2t . , . , A E are n distinct points on a circle that partition the circle into n arcs AjA 2 + AvAz* • . . * A„_iA B> A^Ai that intersect only at their endpoints, then tnAiA 2 + mAvAs + ■ • * + mA n _iA n + ntA Ji A 1 = 360. Proof: Following is Lhe proof for n = 5, The proofs for other values of n arc similar. By repeated application of Theorem 13.15, we get the following: mA L A 2 + mAzA s 4- mAfiA* 4- mA^Aa + mA^A L = mAiA'jAy -f- mAaA4 + 771A4A5 + mA$A l = mAiA^A* 4- rnA+As - mA$A\ = mAiA 2 A 5 -r mAgAi = 360. EXERCISES 13,5 ■ Exercises 1-7 are on the proof of Theorem 13,15. In each exercise, show how to derive the assertion oJ the theorem from the listed equations for the given case. 1. Case 1 5. Case 5 2. Case 2 6. Case 6 3. Case 3 7. Case 7 4. Case 4 8. In the figure, mABC = 240 and mBA'C = 100. Find mAZB and ntAYC. A ^ -^ Z 13.5 Circular Arcs, Arc Measure 583 Exercises 9-20 refer to Figure 13-32. In the figure, P is the center of the circle; A, B, C, are points on the circle; A-P-B-, mBC — 50; and mBD =s 110. no Figure I&32 9, Name five minor arcs determined by points labeled in the figure. 10. Kind tile measure of each of the minor arcs named in Exercise 9. 11. Name the five major arcs corresponding to the five minor arcs named in Exercise 9, 12. Find the measure of each of the major ares named in Exercise 11. 13. Name two semicircles. 14. Which theorem justifies the conclusion thai mCBD = 160? 15. Find m LBFC. 16. Find m L LTD. 17. Copy Figure 13-32 and draw A£, BC t and FC. Find mlBAC, How does m L BAC compare with mBC? ia Show that mlACB = 90. 19. Draw A~D, BD, and PD on your copy of Figure 13-32. Show that ml BAD = [mBlX 20. Find m£ADB. Exercises 21-23 refer to Figure 13-33. In the figure, Pis the center of the circle; A, B, C are points on the circle; and A-P-B. 21. Copy Figure L3-33 and draw FC. Prove thiil ml BAC = \mBC, 22. Prove thai in L ABC = fynAC. 23. Prove that A ABC is a right triangle. Figui* 1*33 584 Circles and Spheres 24. CHAIJ.ENGE PROBLEM. Ill the KgUTe, Pis the center of the circle; A t B, C, D arc points on the circle; and A-F-B. Prove that m£CAD = %mCBD. 13.6 INTERCEPTED ARCS, INSCRIBED ANGLES, ANGLE MEASURE In the discussions that follow we shall he concerned with angles in- scribed in an arc of a circle and about arcs of a circle that are inter- cepted by certain angles. In Figure 13-34a, £ABC is inscribed in ABC and t ABC intercepts AXC In Figure l&34b, £PQR is inscribed in PQR and Z FQR intercepts FYR. Figure 13-34 None of the angles shown in Figure 13-35 is an inscribed angle, but each angle intercepts one or more arcs of a circle. In Figure 13-35a, AAFB intercepts AXB. In Figure 13-35h, Z R ST intercepts RYS, In Figure 13.35c £AVB intercepts AKR and also CMD. In Figure 13-35d, Z GEF intercepts GHF, In Figure 13-35C, Z A VB intercepts ARB and also CSB, In Figure 13-35f t £AVB intercepts Ai5i and also AGB. 13.6 Intercepted Arcs, Inscribed Angles, Angle Measure 585 Figure 13-35 On the other hand. L LKP shown in Figure 13-36 does not intercept an are of the circle. Figure 1U-H0 586 Circlet and Spheres Chapter 13 Our formal definitions of an Inscribed angle and of an intercepted arc follow. Definition 13.15 An angle is said to l>e inscribed in an arc of a circle and is called an inscribed angle if and only if both of the following conditions are satisfied ; 1 , Each side of the angle contains an endpoint of the arc. 2. The vertex of the angle is a point, but not an endpoint, of the arc. Explain why each of the angles shown in Figure 13-35 fails to be an inscribed angle. Explain why each of the angles shown in Figure 13-34 is an inscribed angle. Draw a picture of an angle inscribed in a semicircle. Definition 1X16 An angle is said to intercept an arc of a circle and the arc is called an intercepted arc of the angle if and only if all three of the following conditions are satisfied: 1. Hie end points of the arc lie on the angle. 2. Each side of the angle contains at least one endpoint of the arc. 3. Each point of the arc, except its endpoints, lies in the in- terior of the angle. You should check to see that each of the figures shown in Figure 13-35 satisfies all three of the conditions stated in Definition 13.16. Which of the tlircr- conditions stated in Definition 13.16 are not satis- fied by the figure shown in Figure 13-36? Figure 13-37 shows two angles inscribed in the same arc. (Of course, the angles intercept the same arc) Name the arc in which both angles, LABD and £ ACD, are in- scribed. Name the arc that both of these angles intercept, it appears that Z ABD and LACD in the figure are congruent. (Measure each of them with your protractor,) That they actually are congruent is a cor- ollary of our next theorem. Figure 13-37 13.S Intercepted Arcs, Inscribed Angles, Angle Measure 587 THEOREM 13.16 The measure of an inscribed angle is one-half the measure of its intercepted are. Proof: Let a circle with center V be given and let Z ABC be inscribed in ABC. Then the intercepted arc is AC* We must prove mLABC = JmAC There are three possible cases as suggested in Figure 13-38. Casel. VisonZA^C. Case 2. V is in the interior of Z ABC, Case 3. V is in the exterior of /ABC. Figure 13-38 Proof of Case I: Since V is on Z ABC, then either AB or BC is a di- ameter of the given circle. Suppose BC is a diameter as in Figure I3-3&L Draw VA Then AAVB is isosceles (Why?) and mLA = m it ABC /.AVC is an exterior angle of AAVB, so by Theorem 7-30, m/AVC = mlA + mLABC Since m£A = m£ABC r n-UAVC = mLABC + mZA3C, or But Therefore 2?n£ABC = mlAVC, m£AVC = mAC (Why?) 2mZABC= mAC mLABC = ^niAC. This completes the proof of Case 1 . Proof of Case 2: V is in the interior of /.ABC as shown in Figure 13-38b, so B Vis between BA and BC. Let B' be the point where opp VB intersects the circle; then BW is a diameter and B' is an interior point 5SB Circles irwj Spheres Chapter 13 of the intercepted arc AC. It follows from Case I that m I ABB' = $mAB' and ml B'BC = ^mBC Adding, we get ml ABB' + m/B'BC = \mAB> + ^nB~C, or ml ABB' + mlWBC = J(w*dB' + mB'C). But ml ABB' + ml B'BC = ml ABC (Why?) and mAB' + miPc = wtAC : Why?) Therefore mlABC= fytiAC and the proof of Case 2 is complete. The proof of Case 3 is assigned as an exercise. Recall how congruent segments and congruent angles are defined in terms of their measures. How would you define congruent arcs? Write what you think is a good definition of congruent arcs. Turn back to Figure 13*30. Does your definition allow the conclusion that arcs AXB and CYD of Figure 13-30 arc congruent? If it does, you should reword your definition so that it excludes this conclusion. Compare your definition with the following. Definition 2,127 Two arcs (not necessarily distinct) are congruent if and ony if they have the same measure and are arcs of congruent circles. Notation H arcs AB and CD are congruent, we write AB = CD. Note mat we cannot say that two arcs arc congruent unless we know that they have the same measure and we know that they are in the same circle or that they are in congruent circles. The following three corollaries are important consequences of The- orem 13.16, Their proofs are easy and are assigned as exercises. COROLLARY l&W.l Angles inscribed in the same are are congruent. COROLLARY 13.16.2 An angle inscrilied in a semicircle is a right angle. 13.6 Intercepted Are*, Inscribed Angles, Angle Measure 539 COROLLARY 13.16.3 Congruent angles inscribed in the same circle or in congruent circles intercept congruent arcs. Figure 13-39 shows two distinct purullel lines intersecting a circle. We call an arc whose endpoints arc on the lines, one endpoint on each of the lines and all of whose interior points are between the lines, an intercepted arc. Thus, in Figure 13-39, lines An and CD intercept arcs AXC and BYD. Are these two arcs congruent? They are congruent by our next theorem. Figure 13-39 THEOREM 13.17 If two distinct parallel lines in the plane of a circle intersect that circle, they intercept congruent arcs. Proof: Let distinct parallel lines / and in intersect a given circle in arcs AXC and BYD. There are three possible cases as suggested in Figure 13-40. B B Figure 13-40 Case L Both I and m are secant lines as shown in Figure 13-40a. Case 2. The line if is a tangent line and the other line m is a secant line as shown in Figure 13~40b. Case 3. Both I and m are tangent lines as shown in Figure 13-4Qc, 590 Circles arui Spheres Chapter 13 Proof of Case 1: Since / and m are secant lines, I intersects the circle in distinct points A and B, and m intersects the circle in distinct points C and D. (See Figure 13-40a.) We must prove that AXC m BYD. Draw B€. Then and But Therefore or mlBCD = ml ABC = $mAXC. mlABC = m L BCD. Why? \mAXC s JmBY/3 mAXC = mBYD. AXC B BYD. Why? This completes the proof of Case 1 of Theorem 13,17, The proofs of Cases 2 and 3 are assigned as exercises, EXERCISES 13.0 Exercises 1-17 refer to Figure 13-4L In the figure* P is the center of the circle: A, B, C, D arc points on the circle; AC is a diameter; mAD = 100; and mBC b 40. Kipiw 13-41 1. Kamc four inscribed angles and name the arc each intercepts. 2. Name the five minor arcs determined by points labeled in the figure. 3. Name the five major arcs corresponding to the five minor arcs named in Exercise 2. 4. Name L\vr> semicircles. 5. Find the measure of each of the minor arcs named in Exercise 2, CJ. Find the measure of each of the major ares named in Exercise 3. 13.6 Intercepted Arc*, Inicribed Anglei, Angle Measure 7. Find the measure of each of the angles ZA, Z B, L C t and Z0. 8. Find the measure of £CFD, 9. Find the measure of LABC. 10. Name an angle thai [S congruent to ZB, 11. Name an angle that is congruent to LD, 12. Prove that AD 1 CD. 13. Find the measure of Z DCB. 14. II AC intersects BD at K, name a pair of similar triangles, 15. If AC intersects BD at E, prove that m / DEC = %{mCD + mAU). 16. Prove that AADE - AW(7 17. Prove that DE ■ £# = AE - EC, 591 18. With each chord of a circle which is not a diameter there are two asso- ciated arcs of the circle. One of the arcs is a minor arc and the other arc is a major arc. The endpoints of the chord are the end points of the arcs. Complete the proof of the following theorem. THEOREM 13. IS In the same circle, or in congruent circles, two chords that arc not diameters are congruent if and only if their asso- ciated minor arcs are congruent. Proof: Using the notation of the figure, we are given two congruent circles , C a nd C, with centers P and F, respectively. AB is a chord of C and A'W is a chord of C. There arc two things to prove. (a) If AB ~ A'B'. then A B tt A'B', (b) If AB s j?B', then M a A^F. {Hints In proving (a), show that AAPB s AAT7*' by the S.S.S. Pos- tulate. In proving (b), show that AAPJ? at AA'P'B' by the S.A.S. Postulate.'! 592 Circles and Spheres Chapter 13 19. Does Theorem 13.18 of Exercise 18 still hold if we replace "minor arcs" with "major arcs" in the statement of the theorem? 20. Prove Case 3 of Theorem 13.16. (Refer to Figure 13-38c.) 21. Prove Corollary 13.16.1. 22. Prove Corollary 13.16.2. 23. Prove Corollary 13.16.3. 24. Figure 13-42 shows three cases of a chord AT of a circle with center P and a tangent VTto tlie same circle intersecting at the point of taugency V. Using the notation of the figure, the angle whose sides are rays VT and VA is sometimes called a tangent-chord angle. Complete the proof of the following theorem. THEORtlM 1^.19 The measure of a tangent-chord angle is one-half the measure of its intercepted arc. Proof: Using the notation of Figure 1 3-42, P is the center of the given circle, AXV is the intercepted arc, AT'is the given chord, and VTis the given tangent. There are three cases to consider. Figure 13-48 Case L P is on VA as shown in Figure 13-42u. Case 2. P k in the exterior of I AVT as shown in Figure 13-42b. Case 3. P is in the interior of I A VT as shown in Figure 1 3-42c. Proof of Casn h (See Figure 13-42a,} AV is a diameter; hence mZ AVT = 90 (Why?) and mAXV = 180. Thus ml AVT = fynAXV. Proof of Case 2: (See Figure 13-42b.) Draw diamotcr VB. mlBVl = imHXV, Why? = limM +mAXV), Why? = {niffi + tyiAX$. Since mlBVT = mlBVA +mZ AVT, we have ml BVA + mlAVr = tynBA + intAXV fey the Substitution Property of Equality. Rut 13.6 Intercepted Arc*, inscribed Angles, Angle Measure 593 mlBVA = imBA. Why? Therefore, by the Addition Property of tonality, we get mlAVT=%mAXV, Complete the proof of Theorem 13,19 by proving Case 3, 25. Prove Case 2 of Theorem 13.] 7. 20. Prove Case 3 of Theorem 13.17. {Hint: In Figure 13-40c, prove that W is a diameter. It will then follow that HXD and BYD are semicircles and that BKD S BYD.) 27. A quadrilateral is said to he inscribed In u circle and is called an in- scribed quadrilateral if all of its vertices are on the circle. Prove that the opposite angles of an inscribed quadrilateral are supplementary. 28. If the diagonals of an inscribed quadrilateral are diameters, prove that the quadrilateral is a rectangle, (See Kxerci.se 27.) 29. Prove that the midray of a central angle of a circle bisects the arc inter- cepted by the angle. 30. The figure shows two secants intersecting in an exterior point V of a circle. The rays VA and VC are sometimes called secant-ray?; and the angle whose sides are these rays is called it secant-secant angle. Com- plete the proof of the following theorem. THEOREM 13.20 The measure of a secant-secant angle is one-half the difference of the measures of the intercepted arcs. Using the notation of the hgure t we must prove fliZV = JrimBYD - mAXC), We have mZV+ ml ABC - m L BCD. why? Therefore mZV+ \mAXC = l 2 mBYD. Why? Complete the proof. 594 Circles and Spheres Chapter 13 31. An angle whose sides are a secant-ray and a tangent-ray from an exterior point of a circle is called a secant-tangent angle. Prove that the measure of a secant-tangent angle is one-half the difference of the measures of the intercepted arcs, (Hint: Using the notation of the figure you must prove that mZ V = fynBYC - mAXC). Use Theorem 13.19 stated in Kxercisc 24. Also see Exercise 3CLJ 32. An angle whose sides are two tangent-rays from an exterior point of a circle is called a tangent-tangent angle. Prove that the measure of a tangent-tangent angle is one-half the difference of the measures of the intercepted arcs. (Hint: Using the notation of the figure you must prove, that m£V= limAYB - mAXB) Use Theorem 13.19 stated in Exercise 24. Also sec Exercise 30.) 33. Copy and complete the following theorem which combines the state- ments of Theorem 13.20 (see Exercise 30) and Exercises 31 and 32 into a single statement. 13.6 Intercepted Arc*, Inscribed Angles, Angle Miaiure 595 THEOREM 13.21 Jf an angle has its vertex in the exterior of a circle and if its sides consist of two secant-rays, or a secant-ray and a tungeut- ray, or two tangent-rays to the circle, then Hie measure of the angle 34. Complete the proof of the following theorem. THEOREM 13.22 The measure of an angle whose vertex is in the interior of a circle and whose sides are contained in two secants is one- half the stun of the measures of the intercepted arcs. (An angle such as the one described in Theorem 13.22 is called a chord-chord angle. Using the notation of the figure, it is given that S3 and CD are secants intersecting at E, a point in the interior of the circle. Prove that ml DEB = ^rnDYB + mAXC). Exercises 35-44 refer to Figure 13-13. In the figure, VA and VC are secant- rays of the circle with center P, VE is a tangent-ray, chords A~D and BC intersect at F, and B-P-C. If the measures of arcs AB, BD, £*£, and AC are as shown, use the results of Exercises 30, 31, 32, and 34 and other theorems proved in this section to find the indicated measure in each exercise. 35* m£AVG 36. mCE 37. mlCVE 38. mLAVE 39. mlBFD 40. mlAFB 4L mLDEG 42. ml BEG 43. ml CAB 44. mlBEA Figure 1*43 596 Circlet and Spheres Chapter 13 45, challenge PROBLEM* In the figure, circles C and C with centers P and F t respectively, are tangent internally at R and Circle C contains P. If RH is any chord (with one endpoint atfi) of C and if H? intersects C at Af, prove that M is the midpoint of KS. 13.7 SEGMENTS OF CHORDS, TANGENTS, AND SECANTS If two distinct chords of a circle intersect at an interior point of the circle, the point of intersection together with the midpoints of the chords determine four distinct segments (other than the chords) which are subsets of the chords. For example, in Figure 13-44, chords AB and CD intersect at P. The four segments to which we refer are segments Ai\ PB, t J", and PD. I 1 is eas> to prove that the protlut 1 \>i 1 he lengths of the two segments on AB is equal to the product of the lengths of the two segments on CD. ^- +^ Figure JIM4 THEOREM 13,23 If two chords of a circle intersect, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other. Proof: Using the notation of Figure 13-44, we are given a circle with chords AB and CD intersecting at P. We are to prove that AP-PB = CP-PD, 13-7 Segment* of Chords, Tangents, and Secants We draw AD and EC. Then 597 LA=z LC £Dm LB. (Why?) (Why?) and Therefore AADP - ACBP by the A, A, Similarity Theorem. It follows that (AP, PD) = (CF, PB) and that AP-PB = CP*PD. This completes the proof of Theorem 13,23. Figure 13-45 shows a line VT tangent to a circle ut Tand a secant FA intersecting the circle in points A and B. in die statements of our next three theorems, we refer to segments such as VT, PB, and FA. Figure 13-45 Therefore it is convenient to have a name for each of them. We make the following definition. Definition 13.18 If Vand T are distinct points and if the line VT is tangent to a circle at T, then the segment VT is called a tangent-segment from V to the circle. If secant PA intersects a circle in points A and B such that A is between P and B t then the segment FB is called a secant-segment from P to the circle and the segment 1*A is called an external secant-segment from P to the circle. The proof of our next theorem is Exercise 37 of Exercises 13.3. 598 Circles and Spheres Chapter 13 THEOREM 13.24 The two distinct tangent-segments to a circle with center O from an external point P are congruent and the angle whose vertex is P and whose sides contain the I wo tangent- segments is bisected by the ray PO. Proof: Using the notation of Figure 13-46, given a circle with center O, tangent-segments PA and PB, and ray PO, we are to prove that J5i fiS FE and that PO bisects LAPB. Give reasons for the statements in the proof when asked, Z CMP and Z OBP are right angles. YVhy? Therefore A GAP and AOBP are right triangles, OA^OB :Why?) and FO^FO, Therefore A OAP ^ A OBP. { Why?) It follows that PA Si PB and that ZAPO ^ ZBPO. To complete the proof we need to show that O is in the interior of LAPB. The union of the given circle and its interior is a convex set. Call it S. If Q is any point of S, then OQ < OA, If R is any point of PA, ex- 4---* ccpt A, then OR > OA. Therefore S does not intersect PA except at A. Therefore all of S except A lies on one side of PA. In particular, O lies on the B-side of PA. Similarly, it may be shown that () lu.-s mi the A 'Side of PB. Therefore O is in the interior of Z.APB and since Z.APO « ZjBPO, it follows that PO bisects ZAPB. 17JEOBEM f.3.25 The product of the length of a secai it-segment from a given exterior point of a circle and the length of its external secant-segment is the same for any secant to the given circle from the given exterior point. Figure 13-46 Figure 13-47 13.7 Segments of Chords, Tangents, and Secants 599 Using the notation of Figure 13-47, given secant-segments FE and FD and external secant-segments J5t and JPG, we are to prove that PR • PA = TO ■ PC. Plan: Prove APCB ~ APA D. Proof: Assigned as an exercise. Our next theorem gives us still another relation between the prod- ucts of the lengths of certain segments. Figure 13-48 shows an exterior point P of a given circle and a tangent to the circle at C which contains the exterior point P. If t is any secant to the given circle which contains P and intersects the circle in points A and B, we can prove that the product PA * PB is the same as the product PC ■ PC, or (PC) 2 . Figure 13-*S THEOREM 13.26 Given a tangent-segment PC from P to a circle at C and a secant through P intersecting the given circle in points A and B, then PA • PB = (PC) 2 , Proof: Using the notation of Figure 13-48, we are given a tangent- segment TC from an exterior point P to the circle at C PB is any secant- segment from P and intersecting the given circle in points A and B. We are to prove that PA*PB = (PCp. L mlPCA = ^mAC 2. m/,B = Jmulfi 3. mLPCA = mLB 4. ZPCA = LB 5. ZPs ZP 6. APCA ~ APBC 7. (PC, PA) = (PB, PC) 8. PA*PB = (PQ 2 Benson 1. The measure of a tangent- chord angle is one-half tile measure of the intercepted arc. 2. Why? 3. Why? 4. Why? 5. Why? 6. Why? 7. Why? 8. Why? 600 Circles and Spheres Chapter 13 Tt follows from Theorem 13.23 that if P is any point in the interior of a circle, the product PA • PB remains unchanged for any chord 7lB which contains P. Also, if P is a point in the exterior of a circle as in Theorems 13.25 and 1 3.26, the product PA - PB remains unchanged for any secant-segment from V and intersecting the circle in points A and B, or for any tangent-segment from P to the circle. Theorems 13,25 and 26 suggest that the value of the "unchanged product" is the square of the length of the tangent-segment from P to the circle. This seems to have no significance for Theorem 13.23. Figure 13-49 suggests (with a bit of help from Pythagoras) that (OP) 2 — r 2 exists even when P is inside the circle. As we shall see, the idea of {OP} 2 — r 2 as the value of die "unchanged product*' is what relates Theorem 13,23 to Theorems 13.25 and 13.26. Figure 13-49 Definition 13,19 Given a circle S with center O and radius r y and a point P in the same plane as 5, the power of /' with respect to S is (OF) 2 — r 2 . Given a circle S with center and radius r, a point P coplanar with S is in the interior of S, on S, or in the exterior of S, according to whether OP < r, OP = r, or OP > i\ hence according to whether (OP)'* < r'^iOF} 2 = i* or {OP) z > r 2 , and hence according to whether the power of P with respect to S is negative, zero, or positive. If an iy-coordinate system is set up in the plane of S with its center O as origin, then the power of P with respect to S is (OF)* - r* = x 2 + if - r 2 and interior of S = {(*, y) : x 2 + if - r* < 0} S = {(*, y) : x 2 + y* - r 2 = 0} exterior of S = {(%, y) i x 2 + y 2 — r 2 > 0). Compare this with the rep rose utations using set-builder notation in Figure 13-9. 13.7 Segments of Chords, Tangents, ami Secants 601 We are now ready for the theorem that relates Theorem 13.23 to Theorems 13,25 and 13.26. THEOREM 13.27 Let a circle S with center O and radius r be given. Let P be a point in the plane of 4? and let p be the power of P with respect to S. If a line through P intersects S in points A and B t then 1. PA-PB= -p if Pis inside S, and 2. PA • PB = p if P is on S or outside S. Proof; Our proof is by cases. In Cases 1 and 3, EF is the diameter of S such that EF contains P, Case I. PA • PR = FE • PF (Why?) = (r - Of)(r + OP) = r 2 - (OP) 2 = -P- Casel J» = A Case 2a. PA • PB = ■ PB = = (Of) 2 - r 2 Case 2fr. PA ■ PB = ■ • = = (Oi^ 2 - r* Case 2a Case 2b Cow 2c. PA-PB = PA • ■ = (OP)2 - r2 = p. ^■^ /i Case 2c 602 Circles and Spheres Chapttr 13 Cases 3a and 3b. FA-PB = PE-PE (Why?) = {OP + r)(OP - r) = (Of) 2 - r* O.-iao 3b The theorems wc have proved in this section enable us to do many numerical problems. Example 1 In Figure 13-50. CD = 38, CE = 2Q S and BE ^ 24. Find DE and AE. Solution: DE = CD - CE = 38 - 20 m 18. Let AE = %; then, by Theorem 13.23, *•££ = CE-DE. Thus *-24 = 20-18 or 24* S 360 Figure 1340 and x= 15. What is the power of £ in this example? Is the power of E a positive number or a negative number? Example 2 In Figure 13-51, PB = 70 and PC = 40. Find PA. Figure UMQ 13,7 Sagment* of Chords, Tangents, and Secants 603 Solution: By Theorem 13.26, (PA) 2 = PB • PC. Therefore (PA) 2 = 70-40. Thus and (PA) 2 = 2800, (FAV = 400 ■ 7. FA = V400-7, PA = 20 yT What is the power of P in Example 2? Is the power of P a positive number or a negative number? Exampte 3 In Figure 13-52, FB = 78, AB a 26, and PD = 82. Find CD. D Figure 13-52 Solution: First we need to find PA and PC. Why? We have PA = FB - AS = 78 - 26 = 52, By Theorem 13.25, PD-PC = PR -PA. Therefore 82-PC=78-52 PC = 78-52 m 49 19 82 41 Thus CD = FD - PC a 82 - 49$ = 32|f What is the power of F in Example 3? 604 Circles and Spheres Chapter 13 EXEBCISES 13.7 I, Prove Theorem 13.25. (liefer to Figure 13-47.) Exercises 2-5 refer to Figure 1 3-53 which shows two intersecting chords of a circle. Figure 13-53 2L UCE= 7 and DE = 9, find the power of E. 3. If AB = 24, BE = 8, and C£ = 12, find DE and the power of E. 4. If CE = -r, DE m 8. AE = 12, and BE = x - 2 f find CE f BE, and the power of E. 5. If AE = x BE = x - 4, CE = 4, and ED = 16 - x, find AE, BE, £D, and the power of E. 1 low are A"U and CD related? Exercises 6-14 refer to Figure 13-54 which shows two secant-raj's and a tangent-ray from an exterior point of a circle. If an answer is an irrational number, put it in simplest radical form. (For example, \/32 = 4 \/2 i» sim- plest radical form. ) Figure 1334 6. If PE =16,PD= 10, and PB = B t find PC. 7. Find the power of P in Exercise 6, 8. Find PA in Kxercise 6, 9. If ED = 9, DP = 12, and PC m IS, find BC. 13.7 StgmanU of Chords, Tangents, and Secants 605 10, Kind the power of F in Exercise 9, 11, Find PA in Exercise 9, 12, If PA = 16, FB = 10, and PE = 24, find DE, PC, BC, and the power ofP< 13, If PA m x, PE = 50, PD = 32, and PC = x + 20, find PA, PC, PB, and BC, 14, Find the power of P in Exercise 13, 15, The figure shows two circles intersect- ing at Q and H, a tangent-ray from V to the larger circle at P, a tangent-ray from V to the smaller circle at S. and a lecant-ray from V intersecting the two circles in points Q and B. Prove that VP = VS. (The line QR in the Ggyre is called the radical axis of the two circles. It is the set of all points of like power with respect to both circles.) 16, The figure shows two circles tangent externally at T and vT is a common tangent, A secant-ray from V inter- sects the larger circle in points A and B, and a secant-ray from V intersects the smaller circle in points R and S, Prove VA - VB = VR - VS. 17. Given the figure in Exercise Hi Prove that AVAR ~ AVSB. 18. Given that the sides of quadrilateral ABCD are tangent to a circle at points H, E, /-, P, as shown in the ligurc, prove that AB + CD = AD + BC B 606 Circles and Spheres Chapter 13 19. If a common tangent of two circles docs not intersect the segment join- ing their centers, it is called a common external tangent If it does in- tersect the segment joining, their centers, it is called a common internal tangent. In the figure, Kb is a common external tangent of the two circles with centers P and Q> PR = 21, QS = 6, and PQ = 25. Find AS. (Hint: Draw $T 1 PK at Z) 20. Cive reasons for steps 2 to 8 in the proof of Theorem 13.26. 21. Complete the proof of the following theorem. THEOREM If A and B are distinct points on a circle and if it is any point between A and B, then ti is in the interior of the circle. Proof: Set up an xy-eooiduiate system with the origin at the center of the given circle and with the X-ftXis the perpendicular bisector of AH as shown in the figure. Let r be the radius of the circle. Then there is a number a such that -r<a<r and A& = {(x, y) : x = a). Then the endpoints of AB arc (a, \fF~~- a 2 } and (a, — yr 2 - a 2 }, and a point R{ a t tj) is between A and B if and only if - \/^ - a 2 < y < V^ - <&, if and only if y 2 < r 2 — a 2 . Complete the proof hy showing that OR < r and hence that it is in the interior of the circle. Chapter Summary 607 22. Let line I be a tangent to a circle ul T, Prove that all points of the circle, except 7', are on one side of I in the plane of the circle. {Hint: Let A and B be anv two distinct points of the circle such that A *£'£ mid B ^= 1\ and suppose that A and B are on opposite sides of I. Then there is a point of I between A and B. Why? Therefore I intersects the in- terior of the circle. Why? See Exercise 21. Contradiction?) 23, challenge FROBliKM. Given a right triangle, APQR, with the right angle at H, let C bo the circle with center at P and radius PR as shown in the figure, Then £Jft is a tangent-segment (Why?) and the circle in- tersects QP in two points S and T. Why? Use Theorem 1 3.2fl and prove that {QFf = (QRf + (RF)Z. Thus you will have given another proof of the Pythagorean Theorem. CHAPTER SUMMARY The following terms and phrases were defined in tills chapter. Be sure that yon know the meaning of each of them. CIRCLE SPHERE RADIUS OF CIRCLE (SPHERE) DIAMETER OF CIRCLE (SPHERE) CONCENTRIC CIRCLES ( SPHERES) CHORD OF CIRCLE (SPHERE) TANGENT LINE TO CIRCLE TANGENT CIRCLES TANGENT PLANE TO SPHERE SECANT CONGRUENT CIRCLES (SPHERES) INTERIOR OF CIRCLE (SPHERE) EXTERIOR OF CIRCLE (SPHERE) GREAT CIRCLE CENTRAL ANCLE OF C1RCI £. MINOR ARC OF CIRCLE MAJOR ARC OF CIRCLE SEMICIRCLE DEGREE MEASURE OF ARCS INSCRIBED ANGLE INTERCEPTED ARCS CONGRUENT ARCS TA NGENT— SEGM ENT SECANT— SEGMENT EXTERNAL SECANT— I GMENT POWER OF A POINT 608 Circles and Spheres Chftptor 13 There were 27 numbered theorems in this chapter. You should road them again and study them so that you understand what they say. The fol- lowing are a list of a few of the more important theorems concerning circles. You should know the corresponding theorems with regard to spheres where applicable. THEOREMS 13.3 and 13.4 I ^sl a circle and a line in the same plane be given. The line is tangent to the cirde if and only if it is perpen- dicular to a radius at the outer end of the radius. THEOREM 13.5 A diameter of a circle bisects a chord of the circle other than a diameter if and only if it is perpendicular to the chord. THEOREM 13,8 Chords of congruent circles are congruent if and only if they are equidistant from the centers of the circles, THEOREM Hi. 1ft The measure of an inscribed angle is one-half the measure of its intercepted arc. REVIEW EXERCISES In Exercises 1-22, refer to the circle C* with center P shown in Figure 13-55, Assume that all points in the figure are where they appear to be ia the plane of the circle. Copy the statements, replacing the question marks with words or symbols that best name or describe the indicated parts. Firuit 13-55 1. AB is called a \Tj of *h e circle. 2. PE is called a |T| of the drcle. 3. AT) is called a [TJ of the circle, KB could also be called a (JJcl the circle. 4. AB is called a [T]. 5.IfSnc={T) l then fi? is called a [7] to the circle at '/'. Review Exercises 609 6. If TK n C = {T}, then TK is to FT at I, 7, Those points named in the figure and which arc on the circle are jTJ- S. Those points named in the figure and which are points in the interior of the circle are [?} 9. [?] is a point in the exterior of the circle. 10. L ADF is an [7] angle. 11. L BAD is [?] in arc BAD. 12. Z ADF intercepts arc |Tj. 13. DTE is a \J\ arc of the circle. 14. DAF is a J} arc of the circle. 15. AM} is a [?J. W. mZBAI? is mB7). 17. L EFT is a [T] angle. 18. m/-T=[7J. 19. The power of <? is [TJ. 20. Hie power of K is g$ 22. Tlic power of A is \T\ r 22. A aJ|0 Is similar to AQJ In Exercises 23-29 refer to the sphere S with center Q shown in Figure 13-56, Copy the statements, replacing the question marks with words or symbols that best name or describe the indicated parts. Figure 13-56 23. QR is a [3 of the sphere. 24. If V t Q, Tare collincar, then Vf is a [?] of the sphere. 25. If W, Q, R are coQmear, then the circle with center Q and containing points W t F, R is a of the sphere, 26. RV is a [Tj of the sphere. 27. HP is a [JO- 28. If a n S = {T), then a is Q] to S at T. 29. If a H S = {T}, then Qf is [7] to a at T, 9ft Find the radius of a circle if one of its chords 16 in, long is 6 in. from the center of the circle. 610 Circles and Spheret Chapter 13 31. How far from the center of a circle with radius 16 is a chord whose length is 24? 32. A circle C and a line Hn tin ay-plane are given by C = {{ar, y) : a* + y2 = 36} (a) Write the coordinates of six points that are on the circle. (b) Write the coordinates of three points that are on the line. (c) Write the coordinates of two points that are on the circle and on the line, 33. A sphere S in an xys-coordinate system is given by S = {{*> y, «) : *• + g* + # m 81}. (a) Write the coordinates of eight points that are on the sphere. (b) Write the cooreli nates of two points that are in the interior of the sphere. (c) Write the coordinates of two points that are in the exterior of the sphere. (d) Write the coordinates of two points on the sphere that are endpoints of a diameter of the sphere and are not on any of the three coordi- nate axes. 34. Write an equation of a sphere with center at the origin of an xtp-coor- dinate system and which contains the point (4, —2, 5). What is the radius of this sphere? 35. Let a sphere with radius 10 be given. A segment from the center of the sphere to a chord and perpendicular to the chord has length 4, Find the length of the chord. In Exercises 3o 45 refer to die circle with center P shown in Figure 13-57. Given the notation of the figure and the degree measures labeled in the figure, find the measure asked for in each exercise. You may need to refer to the theorems stated in Exercises 24, 30 t 31. 32, and 34 of Exercises 13,6. Figure 13*5? Review Exercises 611 36. mAB 37. mCT 38. ml BCD 39. nUCPT 40. m£AVC 41. mlCVT 42. mlBTG 43. milBFD 44. mLABD 45. ittilP/V In Exercises 46-50 refer to the circle with center P shown in Figure 13-58. In the figure, VA and VC are secant-rays intersecting the circle in points A* B and C, I ), respectively. VT is a tangent to the circle at 71 Chords AD and BT intersect at E. Figure I3-5S 40, If VA = 12, VB = 20, and VC = 14, find VD and the power of V. 47. If VA = 16, and AB = 9, find VT and the power of V, 48. If AD = 24, AE = 18, and ET m 8, find Brand the power of B. 40, If BE = 12, ET = x, AE = 18, and ED = x - 2, find x, BT, AD, and the power of E. 50. If BE = 9, EA = x, AE = x + 12, and ED = X - 3, find *, AE, £D, and the power of E, Chapter Brooks /Mankmeyar Circumferences and Areas of Circles 14.1 INTRODUCTION In the first part of this chapter, we consider some of the properties of regular polygons that are useful in developing the formula for the circumference of a circle and the formula for the area of a circular re- gion. We usually say "the area of a circle" as an abbreviation for the phrase "the area of a circular region/* or "the area enclosed by a circle." Tti the last part of the chapter, we depart from our formal geometry and present an intuitive approach to the development of the formulas for the circumference of a circle, the area of a circle, the length of an arc, and the area of a sector. We appeal to your intuition in developing these formulas because a formal treatment involves the use of limits, a topic that vou would study in a mathematics subject called "calcu- lus." We use the idea of a limit to make the formulas seem plausible. 614 Circumferences and Areas of Clrdts Chapter 14 14.2 POLYGONS In this section we investigate some of the angle measure properties of convex polygons. We also consider some of the properties of a cer- tain subset of convex polygons called regular polygons. Recall the definition of a convex polygon in Chapter 4. We say that a polygon is a convex polygon if and only if each of its sides lies on the edge of a halfplane which contains all of the polygon except that one side. In this chapter all the polygons with which we are concerned are convex polygons. Therefore, when we speak of a polygon, we mean u convex polygon. Recall that two vertices of u polygon that are endpoints of the same aide are called consecutive vertices, or adjacent vertices. Two sides of a polygon that have a common endpoint are called consecutive sides, or adjacent sides. An angle determined by two adjacent sides of a poly- gon is called an angle of the polygon. Two angles of a polygon are called adjacent angles of the polygon if their vertices are adjacent ver- tices of the polygon. For the polygon ABCDE shown in Figure 14-1, A and B are ad- jacent vertices. AB and AF, are adjacent sides, and Z.A and LB arc adjacent angles of the polygon. Vertices such as A and C, or such as A and D s or such as B and D* and so on, are called ncmadjaceiit vertices of the polygon. A segment whose endpoints are nonadjacent vertices of a polygon is called a diagonal of the polygon. In Figure 14-1, ~KC is a diagonal of the polygon. Name three more diagonals of the polygon ABCDE. How many distinct diag- onals does this polygon have alto- gether? How many distinct diagonals are there that have a given vertex as an endpoint ? In the work I hat follows we are going to be concerned with determining the number of diagonals from an arbitrary vertex of a given polygon. Note that a triangle has no diagonals since each pair of vertices in a triangle is a pair of adjacent vertices. In Chapter 7 we proved that the sum of the measures of the angles of a triangle is 180. We then used this theorem to prove that the sum of the measures of the angles of a convex quadrilateral is 360. i See The- orem 7.33 and its proof.) Let us now review the ideas of this proof W \ 1 1 1 the aid of the new terminology introduced in the study of areas in Chapter 9. 14.2 Polygoni 615 If the quadrilateral ABCD shown in Figure 14-2 is a convex quad- rilateral, then the diagonal AC parti lions the polygonal region ABCD into two triangular regions, ABC and ADC. The union of these two triangular regions is the polygonal region ABCD. In the proof of The- orem 7.33, we showed that the sum of the measures of the four angles of the quadrilateral is the same as the sum of the. measures of a certain set of six angles, three from each of the two triangles. In this way we obtained 2 « 180, or 360, as the sum of the measures of the angles of a convex quadrilateral. Now let us extend this idea to convex polygons of 5, 6, 7, 8, or a sides, where n is a positive integer greater than 4. Figure 14-3 shows pictures of polygons with 5, 6, 7, and 8 sides. The names of these poly- gons are pentagon, hexagon, heptagon, and octagon, respectively. a 616 Circumferences and Areas of Circtat Chapter 14 On a sheet of paper draw four convex polygons, a pentagon, a hex- agon, a heptagon, and an octagon, In each polygon, label one vertex A and draw all distinct diagonals from A. The diagonals partition each polygonal region into a certain number of nonoverlapping triangular regions whose union is the given polygonal region. By a procedure sim- ilar to the one we used with the quadrilateral, find the sum of the meas- ures of the angles of each polygon and summarize the results in a table like the one shown in Figure 14-4, Complete the last two rows of the table on the basis of your computations for the first five rows. The en- tries in the last row shotild l>e formulas involving n. Number of Number of Sum of Measures Sides of Number of Triangular of the Angles Convex Polygon Diagonals from A Regions of the Polygon 4 1 2 2 ■ 180 = 360 5 2 m m 6 3 m m 7 4 m m 8 m m m 9 m m m n m m Figure 14-4 On die basis of the results of the computations suggested, it seems reasonable to conclude that each convex polygonal region of n sides can be partitioned into n — 2 triangular regions by drawing the diag- onals from one vertex. Since die sum of the measures of the angles of each triangle diat bounds a triangular region is 180, the formula (n — 2)180 appears to be a correct formula for determining the sum of the measures of the angles of a convex polygon of n sides Wc slate this result as a theorem. THEOREM 14.1 The sum of the measures of the angles of a con- vex polygon of H sides is (n — 2)180. Proof: Let A be any vertex of the given convex polygon with n sides and let the polygon be A BCD . . . MN as suggested in Figure 14*5. Since a diagonal exists from A to each of the n vertices of the polygon except the vertices A, B, N (Why?), there are n — 3 diagonals from the vertex A. Match A ABC with AC, AACD with AD This es- 14.2 Polygons 617 tablishcs a one-to-one correspondence between the set of n — 3 diago- nals and the set of all triangular regions with the exception of AMN. Figure 14-5 Therefore there are (n — 3) + 1, or n — 2, triangular regions. The union of these n — 2 triangular regions t ABC, ACD, . . . t AMN, is the polygonal region ABCD , . . \l\\ The sum of the measures of all the angles of the triangles that bound these triangular regions is (n — 2)180. It follows from the Angle Meas- ure Addition Theorem that the sum of the measures of all the angles of the triangles AAJ5C, AACO t , > . , AAMN is the same as the sum of the measures of all the angles of the polygon ABCD . . . MN. This completes the proof. An important subset of the set of all convex polygons is Lhc set of polygons whose sides are all congruent and whose angles are all con- gruent. We call such polygons regular polygons. Figure 14-6 shows a regular pentagon ABCDE and a regular hexagon ABCDEF. 618 Circumferences and Areas of Circles Chapter 14 Definition 14 J A regular polygon is a convex polygon ull of whose sides are congruent and all of whose angles are congruent. What do we call a regular polygon of three sides? Of four sides? Note that a rhombus (Figure 14-7a) which is not a square has all of its sides congruent, but it is not a regular polygon. Why? Similarly, a rectangle (Figure 14~7b) which is not a square has all of its angles con- gruent, but it is not a regular polygon. Why? Figure 1.4-7 {a) Rhombus (b) Rectangle Since a polygon of n sides has n vertices and therefore « angles, we have an important corollary of Theorem 14.1 that applies to a regu- lar polygon of n sides. COROLLARY 14.1,1 The measure of each angle of a regular polygon of n sides is (n - 2)180 n Proof: A regular polygon of fl sides has n angles and each of these singles has the same measure as every other angle of the polygon. Since the sum of the measures of the angles is (it — 2)180, it follows that each angle has a measure of - . n Tn Chapter 6 we defined an exterior angle of a triangle. We now extend the definition to convex polygons of more than three sides. Defin ition 1 4.2 Each angle of a convex polygon is called aj i interior angle of the polygon. An angle which forms a linear pair with an interior angle of a convex polygon is called an exterior angle of the polygon. Each exterior angle is said to be adjacent to the interior angle with which it Forms a linear pain 14.2 Polygons 619 Note that there arc two exterior angles at each vertex of a polygon as suggested in Figure 14-8 and that, t>eing vertical angles, they are congruent to each other. It follows that a polygon of n sides has 2n exterior angles. Figure 14-8 Suppose that we are given a convex polygon of n sides and suppose that we choose one of the two exterior angles at each vertex. The chosen exterior angle and the adjacent interior angle of the polygon at that vertex are supplementary. Why? Therefore the sum of their measures is 180, Since there arc n vertices, the sum of the measures of dl the interior angles and the chosen exterior angles is n • 180* But we have shown that the sum of the measures of all the interior angles of the polygon is (n — 2)180. Therefore the sum of the measures of all the chosen exterior angles (one at each vertex) is n(180) - (n - 2)180 = 180« - ISOn + 360 = 360. We have proved the following theorem. TTTEOREM 14.2 The sum of the measures of the exterior angles, one at each vertex, of a convex polygon of n sides is 360. This means that the sum of the measures of the exterior angles of a polygon is independent of the number of sides the polygon may have. Suppose we consider a particular regular polygon, such as a regular hexagon. Wc know that the sum of the measures of all the interior an- gles of the hexagon is (n — 2)180 and that the measure of eacli angle (n - 2)180 it of a regular hexagon is For a hexagon, n = 6. 620 circumferences and Areas of Clrclt* Chapter 14 Therefore the measure of each angle of a regular hexagon is < 6 ~ 2 > 18 ° = 120. 6 It follows that the measure of each exterior angle of a regular hexagon is 60 and that the sum of the measures of the exterior angles, one at each vertex, is 6" ■ 60 = 360, Another way to calculate the measure of each exterior angle of a regular hexagon is to use Theorem 14.2. Since supplements of con- gruent angles are congruent, It follows that all the exterior angles of a regular polygon are congruent. If we choose just one exterior angle at each vertex, there are 6 chosen exterior angles of a regular hexagon and each of them has a measure of I • 360 = 60, This leads us to the following corollary of Theorem \42. COROLTARY 14.2.1 live measure of each exterior angle of a regular polygon of n sides Is ■ . Proof: Assigned as an exercise, EXERCISES 14.2 ■ In Exercises 1-6, use the formula of Corollary i.4.1. 1 to find the measure of each Interior angle of the indicated regular polygon. 1. Pentagon A. Octagon 5, 24-gon (24 sides) 2, Heptagon (7 sides) 4. Decagon (10 sides) 0, ISO-gon 7, Fi nd t he in ensure of each exterior angle of the regular polygons of Exer- cises 1-6 in two differ cut ways, ■ Copy and complete the statements in Exercises 8 and 9 with the word in- creases or the word decreases, 8» As the number of sides of a regular polygon increases, the measure of each interior angle of the polygon [?J. &. As the number of sides of a regular polygon increases, the- measure of each exterior angle of the polygon [£}■ 10. The measure of each interior angle of a regular n-gon is 140. Find n. {Hint: Solve the equation ( " : i. 2 ) 180 _ 14Q f OT ^j 11. The measure of each interior angle of a regular n-gon is 150. Find n. (See Exercise H>.) 14.2 Polygon* 621 12. Can the measure of each interior angle of a regular polygon be 136? Explain. 13. Find the measure of each exterior angle of a regular 12-gon. 14. The measure of each exterior angle of a regular n-gon is 12, Find n. 15. Can the measure of each exterior angle of a regular polygon be 50? Explain. 16. Given a regular hexagon ABCDEE as shown in the figure below at left, prove that ABDF is equilateral. 17. The figure above at right shows a regular pentagon ABCDE whose sides have been extended to form a five-pointed star. Find the measure of each vertex angle of the star (that is, Z P, LQ> LR? L S, L T). 18. If the skies of a regular hexagon were extended to form a six-pointed star, what would be the measure of each vertex angle of the star? 19. Hie sum of the measures of 14 angles of a polygon of 15 sides is 2184. (a) What Is the measure of the remaining angle? ;b) Could the polygon be a regular polygon? (c) Is there enough information to decide whether it is a regular polygon? 20. The sum of the measures of 8 angles of a 9-gon is 1140. (a) What is the measure of the remaining angle? (b) Could the polygon be a regular polygon? Explain. 21. Prove Comllary 14.2.1. 22. Given a regular pentagon ABCDE, prove that diagonal AD is parallel to side EC, 23. Given a regular hexagon ABCDEl', prove that diagonal AD is parallel to side BC. 24. challenge PROBLEM^Given a regular polygon ABCD ... .V of n sides, prove that diagonal AD is parallel to side BC if n > 5. 25. chaij.f.nc;k HKKBJM. Determine the maximum number of acute an- gles a convex polygon can have. 622 Circumferences and Areas of Circle* Chapter 14 14.3 REGULAR POLYGONS AND CIRCLES We biow that three noneol linear points determine exactly one tri- angle. That is> given a set of three noneollinear points there is exactly one triangle which has these three given points as its vertices. We now proceed to show that, given a set of three noncollinear points, there is exactly one circle that contains the three given points. In this sense, we say that three noncollinear points determine a circle. T^t three noncollinear points D, E, F be given as shown in Figure 14-9, Let of be the unique plane that contains D, E, F. In plane a, let 1% be the unique perpendicular bisector of DE and let t-i be the unique perpendicular bisector of FE, Let P be die unique point of intersection of ti and Z 2 - (How do wc know that k intersects Z 2 ?) Then P is equidis- tant from D and E because it lies on the perpendicular bisector of IM> and P is equidistant from E arid F because it lies on the perpendicular bisector of EF. Therefore P is equidistant from D, E, and F. It follows that P is the center of a circle C with radius r = PD = PE = PF. hf-.^ Figure 14-9 Therefore C is a circle which contains D i E x F* Furthermore, C is the only circle which contains D t £, F. For if C were any other circle with center F and radius r* and containing points D, F., F in plane a, then ¥ = FD = FE=FF and F would be equidistant from D, E, and F, Therefore F would lie on h and 1% the perpendicular bisectors of DF and EF in plane a. Since there is only one point that lies on both of these lines, F = P> ¥ = r, and hence C = C. Therefore there is exactly one circle which contains any given set of three noncollinear points. It follows that there is exactly one circle that contains the vertices of any given triangle, We have proved Theorem 14.3. 14.3 Regular Polygons and Circles 623 Definition 14,3 The circle which contains the three vertices of a given triangle is called the circumscribed circle, or cir- cumcircle, of the triangle arid we say that it circumscribes the triangle. The triangle is said to be inscribed in the circle and is called an inscribed triangle of the circle. THEOREM 14.3 A given triangle has exactly one dreumeirde. We now extend the statement of Theorem 14.3 to Include any reg- ular polygon, Wc want to prove that, given any regular polygon, there is exactly one circle that contains all the vertices of the polygon (that is. the given regular polygon has exactly one circumcircle). THEOREM 14.4 A given regular polygon has exactly one circumcircle. Proof: Let a regular polygon of n sides be given. (In Figure 14-10, we have shown a polygon of 6* sides.) Let Q be the unique circle with center P and radius r which contains A, B, and (7. (How do we know this circle exists and is unique?) We shall prove that PD = r and hence that D lies on circle Q, A similar argument could be given to show that each of the verti- ces of the given polygon lies on Q, Figure 14>10 1. PA = PB = PC = r 2. Z2s Z3 3. mLABC = m/BCD 4. mZ2 = m/.$ 5. mZl = mZ4 6. Z 1 == 14 7. AB S DC 8. PB m PC 9. &PBAj= APCD 10, PA B PD 11. PD= PA = r Reason 1. Definition of circumcircle 2. Why? 3. Why? 4. Why? 5. Angle Measure Addition Theorem (3, 4) 6. Why? 7. Why? 8. Why? 9. S.A.S, Postulate {7, 6, 8) 10, Why? 11. Statements JO and 1 Thus D is on Q and this completes the proof for point D, 624 Circumf»r»nc«s *nd Artai of Clrcltt Chapter 14 Definition 14.4 {See Figure 14-11.) The rlmrffimfWllir of a regular polygon is the center of its circumscribed circle. A circumradius of a regular polygon is a segment (or its length) Joining the center of the polygon to one of the vertices of the polygon. An inradius of a regular polygon is a segment ior its length) whose endpoints are the center of the polygon and the foot of the perpendicular from the center of the polygon to a side of the polygon, A central angle of a regular polygon is an angjle whose vertex is at the center of the polygon and whose sides contain adjacent vertices of the polygon. figure I Ml Some authors use radius Instead of circumradius, and apotfiem instead of in radius. We prefer the more descriptive terms* circumradius and hirudins. They are the radius of the circumscribed circle and the radius of die inscril>ed circle of a regular polygon. We know that a regular polygon has a circumscribed circle (Theorem 14.4) and we shall see that a regular polygon has an inscritwd circle (Theorem 14.7). Note that, in connection with a regular polygon, each of the words circumradius and inradius is used in two different ways: ( 1 ) as one of a set of segments and (2) as a positive numl>er. For example, the inradius of a regular polygon means the number that is the length of a segment — any one of the segments defined as an inradius in Definition 14.4. We shall sec in Theorem 14.8 that all these segments have the same length. On the other hand, an inradius of a regular polygon is a seg- ment, usually one of the segments defined as an inradius in Definition 114, but it might also mean any radius of its inscribed circle. Similarly, the circumradius of a regular polygon means a number, whereas a circumradius of a regular polygon usually means any one of the seg- ments joining the center of the polygon to a vertex of the polygon, but it might also refer to any radius of die circuinscril>ed circle. The con- text in which the word is used should make it easy to decide which meaning is intended. It follows from the definition of a central angle that a regular poly- gon of n sides has n central angles. 14.3 Regular Polygons and Circles 625 Suppose that we are given a regular polygon ABCD . . . MN of n sides. (Figure 14-12 shows a regular polygon of 6 sides.) Let F be the circunicentcr of the given polygon. Il follows that all of the triangles, APAB, AFBC, A PCD, , . , , APJVA, are congruent by the S.S.S. Postulate. Kipire 14-12 (Prove that A PAS = A PUG) Let us agree to call each of these tri- angles (that is, a triangle whose vertices are the center and the end- points of a side of the polygon) a central triangle of the regular poly- gon. It follows from the definition of congruent triangles that all of the central angles of a given regular polygon are congruent. We combine these results into the statement of our next theorem. TI7EOREM 14.5 Let a regular polygon of n sides be given. Then all the central triangles of the gjven polygon are congruent and all of the central angles of the given polygon are congruent. Again, suppose that we are given a regular polygon ABCD . . . MN of n sides. (Figure 14-13 shows a regular polygon of 6 sides.) Figure 14-13 Let P be die cii'cumcenter of the given polygon and let PR, PS, FT, » . « » PVbe the n inradii of the polygon. By the definition of an inradius, the 626 Circumferences and Area* 0* Circle* Chapter 14 segments PR, M, PT, . . . , PV are the altitudes to the bases IB, BC, CD, . « < , RA, respectively, of the central triangles, APAB, APBC, APCD, . , . , APNA, of tlic given polygon, By Theorem 14.5 these central tri- angles are all congruent. Since corresponding altitudes of congruent triangles arc congruent, it follows that all the inradii of a given regular polygon are congruent This means that, in the plane of the given poly- gon, the points R, $, T Vile on a circle {) whose center is P and whose radius, a, is the inradius (a number) of the polygon. Since PR I AB^PS^ SC^ff A OR . . . t FT A KE, it follows that each of the sides AB, BC, CD* . . , , NA of the given regular polygon is tangent to the circle Q and that the points R, S, T, , V are their respective points of tangeney. Definition 14.5 A circle is said to be inscribed in a polygon and is called an inscribed circle, or iiicircle, of the polygon if each of the sides of the polygon is tangent to the circle. We also say that the polygon circumscribes the circle. The center of an incircle of a polygon is an incenler of the polygon* We now show that there is exactly one circle inscribed in a given regular polygon. Let a regular polygon such as the one in Figure 14-13 be given. We have demonstrated that the circle Q with center P and radius a is an inscribed circle of the given polygon. It can be shown that a point which is equidistant from the sides of a regular polygon is also equi- distant from its vertices and hence must be the circumeenter. Since there is only one circumeenter, it follows that there is only nine ineen- ter. Therefore Q is the only inscribed circle of the polygon. We have proved the following two theorems. THEOREM 14.6 All the inradii of a given regular polygon are congruent THEOREM 14.7 There is exactly one circle that is inscribed in a given regular polygon. We can now think of the center of a regular polygon as its incenter or its circumcenter. Recall from Chapter 10 that two polygons are similar if there is a correspondence between their vertices such that corresponding angles are congruent and lengths of corresponding sides are proportional. It 14.3 FUgular Polygons and Circles 627 follows from the definition of similar polygons that if two regular polygons are similar, then they have the same number of sides, Is the converse statement true? That is, is it true diat if two regular polygons have the same number of skies, then they are similar? This brings ns to our next theorem, THEORFM 14,8 Two regular polygons are similar if they have the same number of sides. Proof: Let two regular polygons ABCD , . , M .V and A'B'C'D' , . » M'N\ each having n sides, be given as suggested in Fig- Figure 14-14 ure 14-14 We want to prove that the correspondence ABCD . . . MN * — * A'B'Ciy . . . 1TW is a similarity. By the definition of regular polygon and Corollary 14. LI, we have m n and mLA* s m£W = m£C = mlD' = • • • = mlN' _ (n - 2)180 n Therefore mLA = m£A\m£Sm mlB' mlN = mlN', and the corresponding angles of the two polygons are congruent. 628 Circumferences and Areas of Circles Chapter 14 It follows from the definition of regular polygon that there are two positive numbers -9 and $' such that AB = BC-CD = •> =NA = s and AW a B'C = CD' m , , . = N'A' = s'. Then AB = s = -£■•• = 4-A'F, BC = * = 4**' ■ 4-B^ etc. It follows that (AB, BC. CD, , . , , NA) = (A'B\ B'C, CD\ , . ♦ , iVA ) with proportionality constant k = -^ , Therefore the lengths of the cor- responding sides of the two given polygons are proportional and ABCD ...MN~- A'&Cfy . - . M'N'. This completes the proof of Theorem 14,8. THEOREM 14.9 The perimeters of two regular polygons, with the same numl>er of sides, are proportional to the lengths of their circiunradii or their in radii. Proof: Let two regular polygons ABCD . . . MN and A'B'Ciy . . . M f N* t each having n sides, be given as suggested in Figure 14-15. A H B Figure 14-15 Let p and p', s and $', t and r\ a and a' be their respective perimeters., lengths of sides, circumradii, and inradii. Let P and W be the centers of 14,3 Regular Polygons and Circles 629 the two polygons and consider the two central triangles AAPB of ARCD . . , MN at id AA'FB' of A'B'CD' , , . Af.V. Since ml ABB = mlATB' = n (you will be asked to show this in the Exercises), it follows that AAFB — AA'FB' by the S.A.S. Similarity Theorem, Therefore (AB.r,a) = (A r B' f r>,tri. Why? Since corresponding altitudes of similar triangles are proportional to the lengths of any two corresponding sides, we have ($, r, a) = {s\ f, a'}. Finally, p = ns for ABCD . . , MX and p' = ns' for A'B'CD' . . . M'.Y. Since ps' = (ns)s' = (ns> = pfs, it follows that and hence (p. s, r» a) =: (p\ *', /, a'). This completes the proof of Theorem 14.9* Since all central triangles of a given regular polygon of n sides are congruent, and since they partition the given polygonal region Into n nonoverlapping triangular regions, it is easy to prove that the area of a regular polygon is equal to one-half the product of its inradius and perimeter. We state this as our next theorem, TITEORFM 14 JO The area of a regular polygon is equal to one- half the product of its inradius and perimeter, that is, S = -jflp. Froof: Assigned as an exercise. 630 Circumference* and Aft*« vt Clrclei Chapter 14 The last theorem of this section follows easily from Theorems 14.9 and 1410, THEOREM 14,11 The ureas of two regular polygons with the same number of sides are proportional to the squares of their circum radii (or the squares of their in radii, or the squares of their side lengths, or the squares of their perimeters). Proof: Assigned as an exercise. As we said in the introduction to this chapter, the idea of a limit is important in developing formulas for the circumference and the area of a circle. We conclude this section with a brief discussion of se- quences and limits. An infinite sequence of numbers, denoted by {%„}, is a sequence *i» *fc *& • • • , *■» • * « in which, for every positive integer n, *„ is a number. In some applications it is convenient to start counting with some integer other than 1 . Thus £3, jt|, *a, . . . , x n> . , . is an infinite sequence. Example 1 Let the nth term of a sequence be given by x„ = 2n. Write the first five terms and the 40th term of the sequence {2n}, Solution; *! = 2 • 1 = 2. *a = 2*2 = 4. * 3 == 2*3 = 6, x, = 2 ♦ 4 = 8. * 5 = 2-5 = 10. xw = 2 ■ 40 = 80. Therefore die first five terms of the sequence are 2, 4, 6, 8, 10 and the 40th term is 80, Example 2 Let the nth term of a sequence be given by 1* + 1 *» = n Write the first five terms, the J 00th term, and the 1000th term of the sequence m Solution: 1 + 1 *1 = 1 — 2. *2 = 2 + 2 1 — 3 2 *3 = 3 + 1 4 3 3 4 + 1 5 X4 = 4 = 4 14,3 Regular Polygons and Circles 631 5 + 1 6 *S = = — = ~ * 5 5 IftA . 100+1 101 100 " 100 " 100 1000 + 1 1001 * ,000= 1ooo~-iooo In Example 2, as n gets larger and larger, the terms of the sequence gj$f closer and closer to some particular number. What number is it? In Example 1, as n gets larger and larger, do the terms of the sequence {2n} get closer and closer to some particular number? If so, what number is it? If the terms of a sequence {x n } get arbitrarily close (as close as we desire) to a particular number L as n gets larger and larger, we then say that the nth term of the sequence is approaching L (denoted by x n —*L) and we call L the limit of the sequence. Thus, in Example 2 S n and we call 1 the limit of the sequence | 1 , In Example 1, the sequence {2n} has no limit Find the limit (if any) of the sequence — 1 . EXERCISES 14.3 L It follows from Definition 14.4 that a regular polygon of n sides has n central angles. Prove that the measure of each central angle of a regular 360 polygon is — — . 2. Prove that each central triangle of a regular polygon is an isosceles triangle. 3. Prove that each central triangle of a regular hexagon is an equilateral triangle. 4. Prove that the length of a side of a regular hexagon is equal to the circumradius of the circuiucirclc. 632 Circumferences and Area* of Circlt* Chapter 14 5. Prove thai an in radius of a regular polygon bisects a side of the polygon and hence lies on the perpendicular bisector (in the plane of the poly- gon) of the side of the polygon, 6. Prove that the bisector rays of the interior angles of a regular polygon are concurrent at the center of the polygon. 7. Prove that the measure of a central angle of a regular polygon is equal to the measure of an exterior angle of Lhe polygon, 8. Two regular pentagons have sides of length 4 in. and 5 in,,, respectively . What is the ratio of the perimeter of the smaller pentagon to the perimeter of the larger one? What is the ratio of their eircurnradii? Of their inradii? Of their areas? 9. The area of a regular 12-gon is *£■ times the area of a second regular 12-gon. What is the ratio of their perimeters'? Of their eircuinradii? Of their inradii? 10. A regular hexagon has twice the area of another regular hexagon. What is the ratio of the perimeter of the smaller hexagon to the perimeter of the larger one? What is the ratio of the lengths of their sides? Of their cirenmradii? 11. Two regular polygons of the same number of sides have perimeters of 36 in. and 48 in., respectively. The in radius of the first polygon is 3\/5hx What is the hirudin* of the second polygon? What is the area of each of the polygons? 12. Find the circum radius, inradius, and area of an equilateral triangle each of whose sides is of length 2 \/3. 13. Find the circum radius, hirudins, and area of a regular hexagon each of whose sides is of length 12. 14. Show that the inradhis of a regular hexagon is — ^ — . s, where s is the length of a side of the hexagon, 15. Derive a formula for the urea S of a regular hexagon in terms of the length s of its side. (Hint: See Exercise 14.) 16. Use the formula you derived in Exercise 15 to find the area of a regular hexagon each of whose sides is of length 12. Does your answer for die area agree with that of Exercise 13? 17. If a regular hexagon and a regular triangle are inscribed in the same circle, prove that tire length of the side of the Hexagon is twice Lhe ui- radius of the triangle, 18. A square is inscribed in a circle of radius 1, A second square is circum- scribed about the same circle. Find the area and the perimeter of each square, 19. Repeat Exercise 18 using regular hexagons instead of squares. 20. The length of each side erf a regular hexagon is 8 s/3. Find the area of the hexagon in two different ways. 14,3 Regular Polygon* and Circfct 633 21* The circum radius of a regular pentagon is r and the length of each of its sides is s. Find the area of the pentagon in terms of r and s. 22. In Exercise 21, if the eircumradius of a second regular pentagon is 2r, find die area of the second pentagon in terms of t and *. 23. Prove Theorem 14 JO, 24. Prove Theorem 14 J L 25. Write reasons for statements (2), (3), (4), (6), (7), (8), and (10) In the proof of Theorem 14,4. 26. What is the 1000th term of the sequence {-}? What is the millionth term? Which of these two terms is closer to zero? Could we make — as n dose to zero as we desire by choosing n large enough? 27. Find the first five terms, the 100th term, and the 1000th term of the 28. What is the millionth term of the sequence 1 ^ 1 of Exercise 27? Find (he limit (if any} of the sequence. In Exercises 29^37, a formula for x m is given. Find *i, x», x& and Xi<>, What is the limit (if any) of the sequence («■}? ■fc^-l + i M.^iiJOO 1 + 50 ~2n 2u 2 tt 2n n * L *-* + 5T- 2n m *---2nT- 32.^=2- y? tXm = !£jLL 33, ^ = J- 2" 38. Let a circle be given. For each n,n > 3, let a regular polygon of n sides be inscribed in the circle. Let »„ be the perimeter of the n-gon. Does Px, p<» Ps» • ♦ ■ , p«» ■ • • define a sequence? Do you think {p„ } has a limit? If so, how would you describe the limit? 39. 1 ,et a circle be given. For each n, n > 3, let a regular polygon of n sides be inscribed in the circle. I .et S n be the area of the polygon. Does S3. S*. Sa, . , , , S nw . , . define a sequence? Do you think {S„} has a limit? If so, how would yon describe the limit "J^- . — -jT — J has a limit. What is it? 634 Circumferences and Are« of Circle* Chapter 14 14.4 THE CIRCUMFERENCE OF A CIRCLE Thus far, in our formal geometry, length has not been defined for anything except segments. If a path from one point to a second point is such that every point of the path lies on the same segment, then the length of that path is, of course, the length of the segment joining the two points. However, If the path is a circular arc, what is the distance from the first point to the second point along the circular arc, that is, what is the length of the arc? The degree measure of the arc would not be a satisfactory way of describing its length since it is possible for two arcs to have the same degree measure and Lo have different lengths as suggested in Figure 14-16. Figure M-I6 Each of die arcs AB and A B' shown in the figure has a degree measure of 90. But it certainly seems reasonable to think of the arc AB as having a greater length than the arc A B\ Wc start by explaining what we mean by the length of circular arcs and then deriving ways of finding such lengths. We first proceed informally, referring to the physical world. You may have been asked, at one time or another in yonr study of informal geometry, to wrap a string around a circular object, then pull it out straight, and measure its length, liy doing this you are able to arrive at an approximation to the distance around the object However, we cannot describe the process of wrapping a string around a circular object in our formal geometry. We call the '"distance around a circle" the circumference of the circle and denote it by C A more sophisticated approach to finding an approximation to the circumference of a dido is in terms of the perim- eLers of regular polygons inscribed in the circle. There was no difficulty 14.4 Th« arcumftmtc* of • Circle 635 in defining the perimeter of a polygon because the sides of a polygon are segments and each of these segments has a length. But a circle con- tains no segment of a line, and thus we cannot define its circumference (or perimeter) so simply. It seems reasonable lo suppose that if we want to find the circumference of a circle approximately, we can do it by inscribing in the circle a regular polygon with a large number of sides and then measuring or computing the perimeter of the polygon* Given a circle, let »„ be the perimeter of a regular polygon of n sides inscribed in the circle. Then as n gets larger and larger, the number p„ increases, that is, each term in the sequence {p„} ts greater than the preceding term. For example, we can inscribe a square in a circle. By bisecting the central angles of the square we obtain a regular octagon inscribed in the same circle as shown in Figure 14-17. Using the Tri- angle Inequality Theorem, it is easy to show that the perimeter p$ of the regular octagon is greater than the perimeter p 4 of the square. If we continue to bisect the central angles, wc obtain a regular 16-gon, a reg- ular 32-gon, and so on. The perimeters of these regular polygons p 4 . ptt* Pig* P32. . - ■ form an infinite sequence of numbers, and each term in the sequence is greater than the preceding term. Hfttra 14-17 It is shown in the calculus that if a sequence of numl>ers is increas- ing (that is, if each term in the sequence is greater than the preceding term), and if the sequence is bounded (that is, if there is a number that is equal to or greater than any term in the sequence)* then the sequence has a limit. The squcnoe p*» pg, pm, ps*jt • ■ ■ described above in con- nection with the circle shown in Figure 14-17 can be shown to be bounded. In fact, it can be shown that any square that circumscribes the given circle has a perimeter that is greater than any of the terms in the sequence. 636 Cfreumfcrancet and Areas of Circlai Chapter 14 Let a sequence of perimeters of regular polygons of n sides in- scribed in a given circle be denoted by {p«}. Let die limit of this se- quence be denoted by C, that is, Hm p n = C. We are now ready for our formal definition of circumference as die limit of the p, . Definition 14.6 The circumference of a circle is the limit of the sequence of perimeters p a of the inscribed regular pol- ygons (that is, C = lim p n ). Note that we are forced to use limits in defining the circumference of a circle. In order to derive the formulas for the circumference and for the area of a circle* we need a theorem about limits which we state here without proof. We can treat the theorem as a postulate, although it is not a postulate concerning our formal geometry. Rather t it is a postulate concerning the real number system. THEOREM 14.12 Let {**} and {y n } be two sequences of real munhers with nth terms x n and y n , respectively, 1 . If the limit of x« is Lj and the limit of y n is I^ t then the sequence whose nth term is x„y n has a limit, and ton x n y n = Li'Lz. 2. If die limit of *„ is Lj and the limit of y H is L^ J* 0, then the sequence whose nth term is — has a limit and lim — = — -. 3. If x n = y„ for every positive integer n > I and if {*„) and { y„] each has a Limit, then lim x» = lim i/„. 4- If k is a real number and if x n = k t for every n > l f then lim x n — k. Example 1 If lim ^^=A = 2 and lim -^-- = 3, then n n + 2 rnl U,n(fczLa._aL) = 2-3 = 8 \ n n + 2/ \ n n -f- 2/ 3 2 Find lim x n if x?, = 2 for every » > 1. 14.4 Tht Cfrcumferancs of a Circle 637 Solution; If x, { = 2 tor every n > 1, then {ar rt } is a sequence of num- bers whose every term is 2; that is, {a&,} = 2, 2, 2 By part 4 of Theorem 14.12, lim x n = 2, Before we derive a formula for the circumference of a circle, we need to know that the number ~ , where C is the circumference of a a circle and d is its diameter, is the same for all circles, That the num- ber ~ is the same for all circles is a corollary of our next theorem. THEOREM 14.13 If C and C arc the circumferences of any two circles with diameters d and il\ respectively, then (C, d) = (C. <f). Proof: Let K t K' be any two circles with circumferences C. C and radii r, r', respectively, as shown in Figure 14-18. Flf»U-Ifi Let [p H ] he the sequence of perimeters of regular polygons of n sides inscribed in circle K with radius r and let [p^] be the sequence of perimeters of regular polygons of n sides inscribed in circle K* with radius r\ By Theorem 14.9, (p», r> f (pi />. It follows that hence that (1) (Pm <*) = (Pn, <#% where d and ci are the diameters of the circles K and K', respectively. 638 Circumferences and Artai of Ckrd«i ChapUr 14 By Definition 14.6, lim p n = C and Ump^C, It follows from Equation (I) that (2) £L = iL. By part 3 of Theorem 14.12, we get from Equation (2) (3) lim &■ = lim 4- By part 1 of Theorem 14.12, lim £=■ = £, and, since d and d' are the same for all n, it follows from part 4 of The- orem 14.12 that hm- = -. Substituting these last two results in Equation (3), we have that SL-± C d' or that and the proof is complete. COROLLARY 14.13.1 If C and d are the circumference and di- ameter, respectively, of a circle, then the number ^ is the same for a all circles. Proof: Let K, K' be any two circles with circumferences C, C and diameters d, d\ respectively, By Theorem 14.13, we know that (C, d) - «7, d'\ 14.4 The Circumference of a Circle 639 By alternation, we get (C, C) = (d, d'). Therefore -y = ■*£-, This proves that the number -r is the Same for any two circles. Definition 14.7 If C is the ctreum fcrence of a circle and d is its diameter, then the number -4, which is the same for all a circles, is denoted by the Greek letter r. It follows from Corollary 14,131 and Definition 14,7 that C = ml for any circle with circumference C and diameter d. Since d = 2r, where r is the radius of the circle with diameter d t we have C = 2*rr as another formula for the circumference of a circle. It has been proved that it is not a rational number, that is, v cannot be represented by -^ where a and b are integers with b =fi 0. However, h we can approximate tt as closely as we desire by means of rational num- bers. Some of the more common rational number approximations to sr are 3.14, 2 ^, and 3.1 416. It has been shown that ff, to ten decimal places, is 3.1415926535. EXERCISES 14.4 1. Show that |f{ is a closer approximation to w than is 2 ? 2 2. Show that 2 f z: 3,14 to the nearest hundredth. (We read "^" as *'is approximately equal to.") In Exercises 3-10, C, r t and d represent the circumference, radius, and di- ameter, respectively, of a given circle. Express answers in exact form, in terms of it if necessary. 3. If r = 7, find a 4. If C = 83«r, find ft 5. If d =12.5, find C. 640 Circumferences and Ar»n of Circle* Chapter 14 6. If f = 36.4 in., find C. 7. If C = 36.4 in., find r. & If C = 2**, find r. 9, If r = 3ir, find C. 10. If C= 14*26*, find Id 1L In Exercise 6, use v = -4^- and find C to tlie nearest inch. 12* In Kxerdse 6, use % = 3.14 and find C to the nearest inch. Compare your answer with that of Exercise 11. 13. In Exercise 7, use r, = -3^- and find r to the nearest hundredth of an inch. 14. In Exercise 7, use *r = 3.14 and find r to the nearest hundredth of an inch. Compare your answer with that of Exercise 13. 15. Prove that the circumferences of two circles are proportional to their radii, 10. The circumference of one circle b two-thirds the circumference of a second circle. What is the ratio of the radius of Oie first circle to the radius of the second circle? 17. Two circles have radii of 15 and 23, What is the ratio of the circum- ference of the smaller circle to the circumference of the larger deck? What is the ratio of the diameters of die two circles? 18. A lire on a wheel of a car has a diameter of 28 in. 1 f the wheel makes 12 revolutions per second, wliat is the approximate speed of the car in miles per hour? (Use w == 4j^-.) 19. Given the same car as in Exercise 18, how many revolutions per second would the wheel make if the car were traveling 50 miles per hour? (Use tr= 2 7 2 -) 20. A square ABCD fa inscribed in a circle with center E as shown in the figure. If the radius of the circle is 7, find the perimeter of the square. (Hint: Show that &EAB is an isosceles right triangle.) 14.4 The Circumference of a Circle 641 21. A square ABCD is inscribed in a circle with center £, Another square PQHS is circumscribed about the same circle as shown in the figure. If the radius of the given circle is 14 in., find the following: (a) The perimeter of each square to the nearest inch. (b) The area of each square to the nearest square inch, (c) The circumference of the circle to the nearest inch. (Use v = Q) v E B 22. The perimeter of a square is equal to the circumference of a circle. If the radius of the circle is 1, find the area of the square in terms of ?■ 23. Show that if the radius of a circle is increased by 1 unit, the circumfer- ence of the circle is increased by 2*r units. 24. Show that if the circumference of a circle is increased by 1 unit, the radius of the circle is increased by «=r« units. 7 2-ff 25. Assume that the surface of the earth is u sphere. Imagine that a steel band is fit snugly around iho equator (a great circle of the earth";. Sup- pose that one foot is added to the length of the band and that it is raised a uniform amount all the way around the earth. To the nearest inch. how many inches will the new band be above the earth? (See Exercise 24.) 26. Assume that the surface of an orange is a sphere. Imagine that a steel band is placed around a great circle of the orange so that it just fits. Sup- pose that one foot is added to the length of the hui id and that it is raised a uniform amount all the way around the orange. To (lie nearest inch, how many inches will the new bund Ixs above tlic orange? (See Exer- cises 24 and 25.) 642 Circumference* and Ar»« of C4rol»» Chapter 14 Exercises 27-36 refer to Figure 14-19. ABCD is a square inscrilwd in the circle with center P, The bisector rays of the four central angles of square ABCD intersect the circle in points E t h\ G» H t respectively. The polygon AEBFCCDH is a regular octagon inscribed in the circle, and TQRS is a square circumscribed about the circle with points of tangcucy E, F, C, H. The radius of the circle is 8. If an answer to an exercise is an irrational nunt- her, give a rational approximation to the nearest tenth. Figure l+iu 27. Find the length * of a side of square ABCD, 28. Find the perimeter of square ABCD. 29. Find the area of square ABCD, 30. Find the length ** of a side of the regular octagon AEBFCCDH. 31. Find the perimeter of the octagon. 32. Find the area of the octagon. 33. Find the length s" of a side of the square TQRS. M. Find the jwrimeter of square TQRS. 35. Find the area of square TQRS. 36. Find the circumference of the circle. 37. Imagine an infinite number of regular polygons inscribed in a circle with radius r. The first polygon has 3 sides, die second polygon has 4 sides, the third has 5 sides, and so on. For every n, n > 3. let p„, a n , ft* and S, he the perimeter, iriradius. side length, and area, respectively, of the regular inscribed polygon with n sides. What is lim p m ? lim a£ I in i *»? lim S m ? 14.5 Area* of Circle*. Arc Length: 5ectw of a Circle 643 14.5 AREAS OF CIRCLES; ARC LENGTH; SECTOR OF A CIRCLE In Chapter 9 we considered areas of polygonal regions. Recall that a polygonal region is the union of a polygon and its interior. In this sec- tion we are concerned with areas of circular regions. We make the following definition. Definition 14.8 A circular region is the union of a circle and its interior. As we said at the beginning of the chapter, "the area of a circle" is m abbreviation for "the area of a circular region," or for "the area enclosed by a circle." We now proceed to get a formula for the area of a circle. We already have a formula for the area of a regular polygon of n sides which is where a n is the inradius of the polygon, p„ is the perimeter of the polygon, and S„ is the area of the polygon. If P„ is a regular polygon of n sides inscritied in a circle with center Q and radius r t as shown in Figure 14-20 (in the figure, n = 8), we observe that the area of the inscribed polygon is less than the area of the circle. Figure 14-20 644 Circumf irenc** and Aran of Clrcltt Chapter 14 For the expression where n = 3, 4, 5 t . . . , there are three sequences involved; {S n }, {a n ) t and {p n }. t*et us consider each of these sequences separately, 1. The sequence {*>„}. As noted above, for each n, S„ is always less than S, the area of the circle, The difference between S s and _S can be made arbitrarily small by talcing n large enougli. It seems reasonable then to say that lim S» = S. Definilon 14.9 The urea of a circle is the limit of the se- quence of areas of the inscribed regular polygons. Thus lim S„ = S, by definition. 2. The sequence [a„). Since the length of a leg of a right triangle is less than the length of the hypotenuse of the right triangle, we observe that the inradfus <*, is always less than the radius r of the circle for each particular value of n, However, the difference between a H and r can be made arbitrarily small by choosing n laige enough- Thus it seems reasonable to say that lim On = r, and we accept this feet without proof. 3. The sequence {«„}, By the definition of the circumference C of a circle, lim p n = C, Now we have S n = fop*. By parts 3 and 4 of Theorem 14.12, we get (1) lim S„ = ^lim a n p n > 14.5 Atms of gmlM, Are Ungth; Sector of a Circle 645 By definition 14.9, 2 : lira S» = S. By part J of Theorem 14.12, we get (3) lim anpn = lim o„ ■ lim p n But (4) lim (in = r and (5) lim p rt = C. Substituting the results of (2), (3), (4), and (5) in Equation (1), we obtain (6) S = {tC as a formula for the area of a circle. Since C = 2*rr, we get by substitution into Equation (6), S = ^r-2irr or S = as a formula for the area of a circle— a formula that should be familiar to all of you. We now state this result formally as a theorem. THEOREM 14.14 The area S of a circle with radius r is wr 2 , that is, COROLLAR Y 14.14, 1 The areas of two circles are proportional to the squares of their radii. Proof; Assigned as an exercise. We have defined the circumference of a circle to be the limit of the sequence of the perimeters of the inscribed regular polygons. We now proceed to define the length of an arc of a circle as a certain limit. 646 Circumftnncti and Areas of Circfti Chapter 14 Consider an arc AM of a circle with center V as shown in Figure 14-21. For k > 2, let P u Pa, »%..., Ph-l be points on Ait such that each of the & angles ZAVP b ZP i VP 2 , / P*VP 3 , . .. , Z ft., VB has a measure of ~ m^C3. (in Figure 14-21, k = 4.) Figure 14-21 Let A* = AP t + PjJPa + f»p, + . . . + P*iR Tiros, in Figure 14-21, A 4 = APi + Pjft + P^ + P 3 B. If we bisect each of the central angles £AVP U L PiVP 2 LPh-iVB, we obtain k more points Q u fy Q k onAB such their A 2 * = AQ t + QiP x + P^ 2 -f Q, £ p 2 + . . . + Q fc R In Figure 14-21, As = (A0, + Q l P 1 ) + (P^g + £*P 2 ) + (P 2 £ 3 + ^) 3 p 3 ) + (p*q 4 + g^. Now APi + Pi^i > APi, F,p 2 + <?aP 2 > -PiPa, Pap3 4- Ws > PaP* ^Q* + Q 4 B > P3H, Why? 14.5 Arras of Circles; Arc Langth; Sector of • CJrcl* 647 It follows that As > A*. If we continue to bisect the central angles with vertex V, we obtain a sequence of sums A*, As, Aje, . . •» A„ (where n = k, 2k t 4k t &k t ,* .) in which each term in the sequence is greater than the preceding term. Also, this is a bounded sequence. Each number in it is less than the circumference of the circle. There- fore the sequence {A n } has a limit and we define that limit to be the length of arc A B, Definition MAO The length of arc AB (denoted by I A B) is the limit of [A m ] where A„ == APi + P X P 2 + • • - + Pn-iB and where Pi, P> £ F n -i are n - 1 distinct points of A& subtending congruent angles at the center V of the circle containing AH. We now have two types of measure for arcs of a circle: their de- gree measures and their lengths. Definition 14 J I In the same circle or in congruent circles, two arcs arc congruent if and only if they have the same length. Thus, if arcs AB and &W are arcs of congruent circles and if AB 5b A'B\ we have (1) IA3 = IA T &'* Also, from Definition 1347, if AB s X" # B\ we have (2) mAB = mA%. Hence, if A& and A^T are congruent arcs of congnient circles, then (JAB, life) = {mAB, mAW). lu other words, lengths of congruent arcs of congruent circles are pro- portional to their degree measures. This is a trivial assertion, of course. Suppose, for example, that IAB = 100 and mA$ = 20. What we are inserting is that (100, 100) = (20, 20). 648 Circumferences and Ar*i« o» Circtoi Chapter 14 It is also true (not trivial and \vc shall not prove it here) that the lengths of any two arcs of congruent circles are proportional to their degree measures. We state this as our next theorem. THEOREM 14,15 The lengths of arc* of congruent circles are proportional to their degree measures. Thus, if K and K' {as shown in Figure 14-22) are congruent circles and if A& is an arc of K and A'B' is an arc of K\ we have (IAB> U%) = (mAB t mAW). Suppose that A'B' in Figure 14-22 is a semicircle. Then niAW = 180 and LYB' = *r. Why? Let b\B 1ms denoted by L and mAB be denoted by M. It follows from Theorem 14.15 that Therefore and (/,, vr) = (M, 180). 180L = tttM We have proved the following theorem, THEOREM 14,16 The length L of an arc of degree measure M contained in a circle with radius r is uiLWr, that is, \ 1 g!U/ I \180/ 14.5 Area* of Circle*; Arc Untfh; Sactw of a Circle W9 Example 1 Find the length of an arc of a circle with radius 12 if the degree measure of the arc is 60. Solution; Use the formula of Theorem 14.16 with M = 80 and f = 12, Therefore Exam pit- 2 Find the degree measure of an are of a circle with radius 2 if the length of the arc is 4? ■ •i Solution: Solving the formula given in Theorem 14,16 for M, we M _ (J80U (Show this.) to 3 We are given that L = 4?- and that r = 2. Therefore M-J3L&- 120. sr«2 3 Note that if M = 360 in the formula of Theorem 14.16, we get Thus L equals the circumference of the circle as it should. Figure 14-23 shows a portion R of a circular region which is hounded by two radii, EA, FB, and an arc AB of the circle. We call H a sector of the circle. A more precise definition follows. Hgurc 14-23 650 Circumferences and Areas of Circles Chapter 14 Definition 14.12 (See Figure 14-24.) Given a circle of ra- dius r with center P and an arc AB of this circle, the union of all segments T<$ such that Q is a point on arc AB is called a sector. We call AB the arc of the sector and we call r the radius of the sector. Figure 14-24 Suppose that we arc given u sector of a circle with radius r and center V as shown in Figure 14-25. Let AB 1*? the arc of the sector. For n > 2, let P h i*2, - • . P»_i be n - 1 points on SB such that each of the n angles lAVP lt £P 1 VP*... 1 £P m _ l VB has a measure of — • mAB, (In Figure 14-25, n = 4.) Figure 14-25 A ""^T^J" i y\ r \p* 14.5 Are** of Cbcitft; Arc Ungth; Sector of a Orcle 651 Let (1) An = APi + Fa?* + ■ • + Pn-iB. Let S., be the area of the polygonal region VAP1P2 • •*!* l®l th en (2) Sn = 4fl a (AJ>i) + J*p!A) + • ; ■ 4- i««(P n iB), where a n is the altitude to each of the bases 3JV F^T\ P*-iR ©f triangles AAVPi, AP1VP2, • • • , AP n _iV7J, respectively. From Equa- tion (2) we get (3) % = ^niAPi + PjPa + ■ .. + P„-iR); hence (4) S H = fynA*. by substitution from Equation (1) into Equation (3). We can make the difference between S„ and S (the urea of the sector) arbitrarily small by choosing n large enough. Similarly, the dif- ference between a n and r can be made arbitrarily small by choosing n large enough. Tt seems reasonable, then, that Urn S n = S and lim o„ = r. By Definition 14.10, lim A n = L, the length of arc Afr By applying parts 3 and 4 of Theorem 14.12 to Equation {4) t we obtain (5) lim S„ = Jiim 0*3,, From equation (5), it follows that S = \tL is a formula for die area of a sector with radius r and arc length h, We state this result as our next theorem. THEOREM 14. J 7 The area S of a sector is one-half the product of its radius t and the length L of its arc, that is, S = |«L If we combine the results of Theorems 14.16 and 14.17, we get the following theorem. THEOREM 14,18 If M is the degree measure of the arc of a sector with radius r, then the area «S of the sector is U==s } aw®* that is, \,jou/ Proof: Assigned as an exercise. €52 Circumferences and Areas of Circlet Chapter 14 Example 3 Find the area of a sector with radius 10 if the degree meas- ure of the arc of die sector is 72, Solution: If we use the formula of Theorem 14.1 8 with r = 10 and M = 72, we obtain Note that, in the formula of Theorem 14.17, if L is the circumfer- ence C of the circle, then L = %<nr in id S = ^r • 2frr = wr 2 , as it should for a circle, Similarly, in the formula of Theorem 14.18, if the given arc is the circumference of the circle, then M = 360 and S = (Jlftn* = ««, as it should for a circle. EXERCISES 14.5 ■ In working title exercises of this set, do not use an approximation for w un- less instructed to do SO. L Prove Corollary 14.14.1. Let Sj and Sv be the areas of two circles with radii r t and r%, respectively. Prove that A. Si) ? fe*, *& 2. The area of one circle is \ times the area of a second circle. What is the ratio of the radius of the first circle to the radius of the second circle? 3. The radii of two circles are r and 2r, How does the area of the larger circle compare with the urea of the smaller circle? How does the cir- cumference of the larger circle compare with the circumference of the smaller circle? 4. Show that if r < 2, the area of a circle is less than the circumference of the circle. (Of course, the urea units and the length units are in different systems. If the length units are inches and the area units arc square inches, and if r < 2, then the statement to be proved asserts that the number of square inches in the area is less than the number of inches in the circumference.) 14,5 Aran of Circl**: Arc Ltngth; Stctor of a Clrcla 653 In Exercises 5-11, r is the radius of a circle and S is its ana. 5. If r = 8, find & 9. If r = y/4\, find S. 6. If S = \mv, find r 10. If r = x/TTir. find S, 7. If S = 154, find r. U. If S = 75.30, find r. 8. If r = 3ir, find S. 12. In Exercise 7, use ir - If- and find r, 13. In Excrdsc 9, use ?r = 3.14 and find S to the nearest tenth. 14. hi Exercise 11. use w = 3.14 and find r to tlie nearest tenth. In Exercises 15-18, the area of a drab is given. In each exercise, find die circumference of the given circle, 15. 81*r 17. 81 16. 49w 18. 49 In Exercises 19-22, the circumference of a circle Is gfven. In each exercise, find the area of the given circle. 19. 40*r 21. 40 20. lfor 22. 16 23. The figure shows two concentric circles with radii 7 cm, and 10 cm. The union of the two circles and the shaded portion between them is some- times called an annul us. Find the area of the annulus. 24. A square and a circle have the same area. (a) Find the perimeter of tlie square in terms of the radius r of the circle (b) Find the circumference of the circle In term* of the length i of the side of the square. 25. Find the ureas of the inscribed and curcuinscribed circles of a square with side length 12. 26. In Exercise 25, find die area of the annulus {see Exercise 23) between the two circles. 654 Circumferences and Areas of Circle* Chapter 14 27. The radius of the circle with center C is 24. The radius of the circle with center C is 12, If mX3=60 and mA'fi'=60 T show lAB^ZlAll'. .A 28. For the circles of Exercise 27, if IAB = lA'B' and mAB = 60, find mA'B'. Given a circle with radius r and an arc AB of the circle, use the given infor- mation in Exercises 29-35 to find the indicated measure. 29. If r = 12 and mAB = 45. find IAB, 30. If r = 1 and IAB = ^- t find mAB. 4 31. If /AB = 11^ and mAJI = 60, find r. 6 32. If LAB = ^L and mA3 = 270, find r. 33. If r = 18 and mAB = 240, find /AB. 34. Kr=9 and mAB = 132, find IAB, 35. If r = 4 and fAH = Git, find iriAB. 36. Given a circle witli radius fa 1, find the degree measure of each of the following arcs of the circle. a) AB tf IAB = £ ' 6 (b) .4C if IA 1; = £ 4 (c) ADiflAD = Z (d) AE if iH = J (e) AF if 1AF = |£ 3 t AC if 1AC - $f 4 (g) ah if zaT/ = ^ 6 (h) A/if 7A/ = ff (i) A/ifJAJ = ~ o-- ;j) AK if lAK = ^ (k) Af,ifZAL = 4c AM if 1AM = (m) AJV if IAN = 2f (n) /IP if LA? = ^ (o) AQ if &® = ll£ 14,5 Areas of Clrct««i Arc Length; Sector of a Circle 65S 37. Prove Theorem 14,18, (Hint: Combine the results of Theorems J 4, 16 and 14.17.) 38. The radius of a circle is 12 in, Find the area of a sector with the follow- ing arc length: (ft) 4r; (b) 2.7* (c) 8w (d) -?- 39. The radius of a circle is 15 in. Find the area of a sector with an arc whose degree measure is (a) 60 (b) 144 (c) 1 (d) 330 40. Let a circle with center P and radius r be given. Points A and B arc points of the circle such thai m Z A PB = 120 and the area of sector A PB is 12ir. Find r and /AB. 41* In an xy-plaue, let sets C and I be defined us follows: C = {(.t-, y) : x* + t/ a ■ 64}, I = {{x, u):ij = x). Let P be the point where / intersects C in the first quadrant, let O be the origin of the given xy-plane, and let B he the point cm the positive uc-axis where C intersects the x-axis. Find the area of the sector POB. Exercises 42-44 refer to Figure 14-26 which shows a circle with center V* and chord AB. The shaded portion hounded by the chord AB and the arc AB is called a segment of the circle. l.et h he llw altitude to AB of A AVB, let AB = s t let VA = r = VB, and let L - 1A&. Figure 14-26 42. challenge problem. Derive a formula for h in terms of r and .?. 43. challenge problem. Derive a formula for the area S of the segment of the circle in terms of r, $ t and L 44. See Exercise 43. Find S if r = 5, s = fl, and L = ~. (Use -n = 3.14 and round your answer to the nearest tenth.) 656 Circumferences and Areas of Circlet Chapter 14 45, challenge problkm. The figure shows three congruent circles with centers P, Q, R. Each of the circles is tangent to the other two and the points of tangency are A, B„ C, (a) Show llutt AP{)R is equilateral. (b) If the radius of each of the circles is 12, find the area S of the shaded region bounded by arcs AB t BC, and AC. (Use it = 3.14 and round your answer to the nearest tenth.) CHAPTER SUMMARY In this chapter we defined the following terms and phrases. Be sure that you know the meaning of each of them. INTERIOR ANGLE OF A POLYGON EXTERIOR ANGLE OF A POLYGON REGULAR POLYGON ( JrU I \IS( H IRED CIRCLE (CIRCUMCLRCLE) INSCRIBED POLYGON INSCRIBED CIRCLE (INCIRCLE) CIRCUMSCRIBED POLYGON CENTER OF REGULAR POLYGON CIRCU M RADIUS OF REGULAR POLYGON INRAD1US OF REGULAR POLYGON CENTRAL ANCLE OF REGULAH POLYGON CENTRAL TRIANGLE OF REGULAR POLYGON CIRCUMFERENCE OF CIRCLE THE NUMBER v CIRCULAR REGION AREA OF CIRCLE LENGTHS OF ARC SECTOR ARC OF A SECTOR RADIUS OF A SECTOR Chapttr Summary 657 There were 18 theorems in tliis chapter, most of which consisted of formulas. In developing many of these formulas, we used the idea of a LIMIT of a sequence of numbers. A complete treatment of limits is too difficult for a first course in geometry. Our goal was to give an intuitive feeling for the concept <jf a limit and to make the formulas seem plausible, Be sure that you know the following list of formulas and that you know how- to apply them. Sum S of the measures oftiie interior angles of a convex polygon ofn sides: S s (n - 2)180, Measure m of each angle of a regular polygon of n sides: , , ft - 2 > 18Q , n Sum S of the measures of the exterior angles, one at eacli vertex, of a convex polygon of n sides: S = 360, Measure m of each exterior angle of a regular polygon of n sides: ^360 ft Perimeter of a regular polygon of n sides: p = 11% where p is tlw perimeter and s is the length of a side of the polygon. Area of a regular polygon of n sides: S = iap, whew S is Hie area, a is the inradhis, and p is the perimeter of the polygon. Circumference of a circle: C = vdorC = 2m, where V is the circumference, r is tlte radius, and d it the diameter of the circle. Area of a circle: where S is the area and r is the radius of the circle. Length of an arc of a circle: where L is the length of tlw arc, M i* the degree measure of the arc, and r is the radius of the circle. Area of a sector: S = ±rL, where Sis the area, r is the radius, and L is the length of die arc of the sector. ess Circumferences and Area* of Clrcln Chapter 14 REVIEW EXERCISES 1. Find the measure of each interior angle of a regular pentagon; of a regular 7-gon. 2. Find the measure of each exterior angle of a regular 15-gon. 3. If the measure of each interior angle of a regular n-gon fa 150, find n. 4. If the measure of each exterior angle of a regular n-gon ii 40 t find n. 5* If the measure of each of the n central angles of a regular n-gon is 20, find n. 6- Using only a pair of compasses for drawing circles and a straightedge for drawing lines, explain how yon would construct a regular hexagon; an equilateral triangle? a square. 7. Find the perimeter and the area of a regular hexagon each of whose sides is 20 cm, long. In Kxercises 8-13, the radius of a circle is given. In each exercise, find the area and the circumference of the circle. Use tt = 3.14 and express each answer to the nearest tenth, 8. r = 13.5 cm. 9, r= 1 10, r = 6.04 in. 11. r= 3,14 ft, 12. r = 12 13, r = 62.8 In Exercises 14-19, the circumference Oof a circle or the area S of a circle is given. In each exercise, find the radius of the circle. Give an exael answer in each case. 14. S = 64w 15. C = 64* 16. S = 216V sq. cm. 17. S = 225 sq. in. 18. C = 14&7 ft 19. C = 72 cm. In Exercises £0-24, r is the radius of a circle, L is the length of an arc of the circle, and M is the degree uieusure of the arc. 20. If r = 16 and M = 80, find L. 21, If r = 4 and L = 3ir t find hi Rtvlvw Extrclttt 659 22. If L = -^ and M = 240. find r. 23. If r = 10 and L = i^, find M. 24. ff r a 45 and M m -220, find L. 25. An annulus (the shaded region shown in the figure) has an inner radius of x and an outer radius of x + 3, If its area is 48?r. find its inner and outer radii. Iii the figure, AC?, BC. fixU, and ffiB are semicircles, with A, U. C collinear. If AC = BC = 7, find the area of the shaded region. fa..-?.) £7. Find the area of a sector if the radius is 14 and if the length of the arc of the sector is lit;. 660 Circumferences and Areas of Circles Chapter 14 28, In the figure, /\ABC is ft right triangle with the right angle at C. An, BV, and AC arc semicircles wilh diameters AB t BC : and AC", respec- tively. If a, b, c arc the lengths of the sides EC, KTL AB, respectively, show that the area of the semicircle with diameter c is equal to the sum of the areas of the two semicircles with diameters a and b* 29. In the figure, A ABC is a right triangle with the right angje at C, KB is a diameter of circle JfC, and AKC and BYC arc semicircles with diam- eters AC and BC. respectively. Show that the sum of the areas of the two shaded regions is equal to the area of the triangle- Review Exercises 661 30. CBHJUJKN&B raoBX.EM. In the figure, A BCD is a square euch of whose sides is 10 cm. long, P, Q t R t S are the midpoints of sides A~R, fi€, CD, DA, respectively. Arcs EF, FC, Oil, HE are arcs of circles with centers P, Q, R, S, respectively, and are tangent to the diagonals of square ABCD at points E, P, G, If. (a) Prove that PQRS is a square. (b) Find die area of the shaded region bounded by the four arcs EF, FG t Gtf , and HE. Vim Bucher/Fhoto Researchers Areas and Volumes of Solids 15.1 INTRODUCTION In your Study of informal geometry you may have learned formulas for finding surface areas and volumes of some of the familiar solids. In this chapter we review and extend this phase of geometry- We con- tinue our somewhat informal development of geometry from Chapters 13 and 14. A formal development of these formulas belongs properly in a calculus course. The development of this chapter is designed to make the formulas plausible. Emphasis is placed on understanding the for- mulas and on using them. In Chapter 13 we studied circles and spheres, A sphere may be thought of as a solid or as a surface. In this book a sphere is a surface, and tlie union of a sphere and its interior is a spherical region. In- formally speaking, the area of a sphere is a number that expresses the measure of the sphere, and the volume of a sphere is a number that ex- presses the measure of the associated spherical region. Although prisms, pyramids, cylinders, and cones may be thought of as either solids or surfaces, in this book we consider them as solids. The 664 Areas and Volumes of Solids Chapter 15 area of a cylinder (more properly., the surface area of a cylinder), for example, is a number that expresses the measure of the surface of the cylinder, and the volume of a cylinder is a number that expresses the measure of the cylinder itself, 15.2 PRISMS Figure 15-1 shows some diagrams of prisms. We might think of a prism as the solid swept out by a polygonal region moving parallel to itself from one position to anodier. Each point P in the region moves along a segment PF as suggested in Figure 15-2, and all of these seg- ments are parallel to each other. The prism is the union of all such seg- ments. Wc make these ideas formal with the following definition. Figure 15-1 15.2 Prisms .-■53 Definition 15.1 ;'See Figure 15-2.) Let a and ft be distinct parallel planes. Let Q and Q* be points in a and ft, respec- tively. Let R be a polygonal region in a. For each point P in R let F be the point in ft such that PF || QQ*. The union of all such segments PF is a prism, if QQ' is perpendicular to a and ft, die prism is a right prism. Figure 15-4 In this chapter we shall limit our discussion of prisms to those prisms whose bases are convex polygonal regions, that is f regions whose boundaries are convex polygons* Definition 15,2 (See Figure 15-2.) Let R r be the polygonal region consisting of all the points F in ft. The polygonal re- gions R and R' arc called the bases of the prism, Depending upon the orientation of the prism it is sometimes convenient to call one of the bases the lower base and the other base the upper base. Sometimes we call the lower base simply the base. A segment that is perpendicular to both a and ft and with its endpoints in these planes is an altitude of the prism. Sometimes the length of an altitude is called the alti- tude of the prism. Prisms are often classified according to their bases. Thus a triangu- lar prism is one whose base is a triangular region; a rectangular prism is one whose base is a rectangular region, and so on. 65ft Ar#av and Volumes of Solids Chapter 15 Figure 15-3 shows a triangular prism. The bases are the triangular regions ABC and A'B'C. The triangular region DEF in the figure, which lies in a plane parallel to the plane of the base, is called a crass section of the prism. \- Figure 15-3 Definition 15.3 If a plane parallel to the plane of the base of a prism intersects the prism, the intersection is en' led a cross section of the prism. Is the triangular region ABC a cross section of the prism shown in Figure 15-3? Is the region A'B'C a cross section of the prism? We say that a point P of the lower base corresponds to a point R of a cross section (other than the lower base) if PR , Qty. We call A DEF in Figure 15-3 the boundary of the triangular region DEF. Similarly. A ABC is the boundary of the lower base of the prism, & A'B'C is the boundary of the upper base, and so on. THEOREM I5>] The boundary of each cross section of a tri- angular prism is congruent to the boundary of the base of die prism. Proof: Let the triangular region ABC in plane a be the lower base of the prism and the triangular region A'B'C in plane j8 be the upper base as shown in Figure 15-3. Let y be a plane parallel to a and intersecting AA', BB' t CO in points D % E, F, respectively. Then, by definition, the region DEF is a cross section of the prism. That the region DEF is a 15.2 Prism* 667 triangular region can be proved using separation properties. We omit the details here. We shall prove ADEF « A ABC. If y = a, then D = A, E ! = H. F = C, and ADEF = AABC. Therefore ADEF s AAiiC. Why? Suppose, then, that y=£aas sug- gested In Figure !5-3. Bv the definition of a prism, there are points Q in a and Q'Jn p such that A A' \ QQ' and BB' \ QQ'. Then AA' | BB' and AD || BE. Also, by Theorem 8.12, Aft I DE. Therefore ABED is a parallelogram and 0E == AB. In the same way we can prove W z. AC and FE Sk CB. Therefore A DEF s AABC by the S. S. S. Postulate. Since y is an arbitrary plane parallel to a and intersecting the prism, we have proved that the boundary of each cross section of a triangular prism is congruent to the boundary of its base, and the proof is complete. Since the upper base of a prism is a cross section of the prism, we have the following corollary. COROLLARY 15.1.1 The boundaries of the upper and lower bases of a triangular prism are congruent THEOREM 15.2 (The Prism Crow Section Theorem) /Ml cross sections of a prism have the same area, Proof: IM the prism as shown in Figure 15-4 be given. (We have shown a prism whose base is a polygonal region consisting of five sides. The following argument can be modified to apply to a prism whose base is a polygonal region consisting of n sides, where n > 3.) Let It be the base and let R' be a cross section of the prism. Then R can be divided into nonover- lapping triangular regions 6), £2, *3 as shown in Figure 15-4, Let t' lt & g be the corre- sponding triangular regions in B\ Then the boundary of h is congruent to the boundary of fj, the boundary of h is congruent to the boundary of t 2 , and the boundary of 1 3 is con- gruent to the boundary of £3, Why? The areas of t[, &, t'3 are equal, respectively, to the areas 0* h, h, k- Why? The area of R f is the sum of the areas of r[, if, tk, and the area of R is the sum of the areas of t-u tz, h- Why? Since these two sums are equal, it follows that the area of R' is the same as the area of R. Since R' is an arbitrary cross section of the prism, it follows that all cross sections have the same area and die proof is complete. 668 Areas and Volumes of Solids Chapter IS COHOLLARY 15.2.1, The two bases of a prism have die same an. -a, Proof: Assigned as an exercise. Definition 15.4 (See Figure 15-5.) A lateral edge of a prism is a segment AA'. where A is a vertex of the base and A' Is the corresponding vertex of the upper base. Given any side of one base of a prism, the lateral face of the prism correspond- ing to that side is the union of all segments FF parallel to a lateral edge and with P on the given side of die base. The lateral surface of a prism is die union of its lateral faces. The total surface of a prism is the union of its lateral surface and its bases. Hgu» 15-5 For tlie prism shown in Figure 15-5, AA' is a lateral edge and AHB'A' is a lateral face. Name four other lateral edges and four other lateral faces in the figure. Note that die lateral edges of a prism are parallel to the segment QQ' in the definition of a prism. The segment AB in Figure 15-5 is not a lateral edge of the prism although AB is an edge of the prism. Indeed, it is an edge, or side, of a base of the prism. THEOREM 15,3 The lateral faces of a prism are parallelogram regions and the lateral faces of a right prism are rectangular regions. Fmvfi (See Figure 15-5.) A complete proof involves a discussion of separation properties which we omit here. Suppose that we are given a 15.2 Prisms 669 prism with points labeled as in Figure ,15-5. We shall content ourselves with proving that ABB'A' is a parallelogram and that ABB' A' is a rec- tangle if the prism is a right prism. Similar arguments could he given for each of the lateral faces of the prism. By the definition of a prism, there is a segment (5p' such that AA' || QQ' and M'|| QQ' Therefore Id' \\ BB'. Why? By the defini- tion of a prism, the planes containing the bases are parallel. Therefore it follows from Theorem 8.12 that AH' || AB. This proves that i\BB'A' is a parallelogram. If die prism is a right prism, then ^?' is perpendicular to plane a. Why? Therefore AA'' a (Why?) and AA"' I AB. Why? It follows that ABB' A' is a rectangle. Definition 15.5 A parallelepiped is a prism whose base is a parallelogram region. A rectangular parallelepiped is a right prism whose base is a rectangular region. A cube is a rec- tangular parallelepiped all of whose edges are eongruenL A diagonal of a parallelepiped is a segment joining any two of its vertices which are not contained in the same lateral face or tiase of the parallelepiped. Figure 15-6 shows pictures of a parallelepiped, a rectangular par- allelepiped, and a cube* In each picture, the segment rlS is a diagonal of the parallelepiped. How many diagonals does a parallelepiped have? .'.,!„ Rrc'-juifiuUr Parallelepiped purillatfpspftd an" *■ -' X 1 II _J \ Figure lo-fi Definition 15.6 The lateral surface area of a prism is the sum of the areas of its lateral faces. The total surface area of a prism is the sum of die lateral surface area and the areas of die two bases. 670 Areas and Volumes of Solids Chapter 15 EXERCISES 15.2 1. Copy and complete; All the faces of a parallelepiped (lateral, upper base, and lower base) are [?] regions. 2. Copy and complete: Alt the faces of a [JJ parallelepiped are rectangular regions, 3. Copy and complete: All the faces of a [JJ are square regions. Exercises 4-16 refer to the prism shown in Figure 15-7. In this figure. Pis a point in a and V is a point in /? such thai. PF JL a. In each exercise, com- plete the statement. 4. The region ABCDE is called a [JJ of die prism. 5. The region A'lfCtXE' h caBod u [JJ of the prism. G. BB' is called a (an) |JJ of the prism. 7. There are [Jj lateral edges in all. 8. Counting the lateral edges and the edges (sides) of the two bases t there are [jj edges in all. 9. The parallelogram region BCCB' is called a (art) [JJ of the prism, 10. There are [JJ lateral faces in all. 15,2 Print* 671 It, Counting the lateral faces and the two liases, there arc [T| feces in all 12. A is called a (an) [T) of the prism. 13. There are (TJ vertices in all- 14. If V is the number of vertices, £ is the numher of edges, and F is the number of faces, then V — £ -j- F = [T], 15. PP is called an |T| of the prism. 16. If M' _L 0. then the prism is called a {?}. 17. Prove that the total surface area of a cube is 6e 2 , where e is the length of one of its edges. IS. Fmd the total surface area of a cube whose edge is 7 cm. in length. 19. Prove that the total surface area of a rectangular parallelepiped is lab + 2lic + 2uc, where a and h are the dimensions of the base and c is the altitude of the prism. Draw an appropriate figure, 20. Find the total surface area of a rectangular parallelepiped if the dimen- sions of the base are 4 cm. by 6" an. and if the altitude of the prism is 8 cm, 21. Given that the pentagonal prism of Figure 15-7 is a right prism, that the lengths of the edges of the base are 3, 7, 4, 9 1 and 6, and that the alti- tude is S, find the lateral surface area of die prism. 22. If S is the lateral surface area, a is the altitude, and p is the perimeter of the base of a right prism, prove that S = ap. 23. Use the formula in Exercise 22 to find the lateral surface area of tiia prism in Exercise 21. Does your answer agree with the one obtained in Exercise 2 1 ? 24. Find the altitude of a right prism if the lateral surface area is 336 and the perimeter of the base is 28. 25. Find the perimeter of the base of a right prism if the lateral surface area is 351 and the altitude is 13^. 26. If the base of the right prism in Fxercise 25 is a square region, find the length of each of its edges. 27* Find the total surface area of a right triangular prism if the boundary of each base is an equilateral triangle whose sides have length 10 and If the altitude of the prism is 12. 2S. Ilie area of a cross section of a prism is 32. The lateral surface area is 128- Find the total surface area of the prism. 29. If XB and FQ are two lateral edges of a prism, prove that AB and TQ arc eoplanar. 672 Areas »r»d Volumes of Solids Ctiapler 15 30. Prove Corollary 15.2 J, 31 , Given the rectangular parol lelepiped shown in the figure with AB = 12, BC = 6, and CD = 8, find the length of die diagonal Ail, (Hint: Draw At?. What kind of triangle is A ABC? What kind of triangle is AACD?) 32. In the figure RS is a diagonal of a cube and the length of each edge of the cube is 8. Prove that RS = 8^/5". 33. Prove that the length of every diagonal of a tube is e y/S, where e is the length of one of its edges. 34. challenge problem. Use the Distance Formula for a three-dimen- sional coordinate system to prove that the diagonals of a rectangular parallelepiped have equal Lengths. 35. challenge PROBLEM. If h is the altitude of a prism, prove that h <, r, where r is the length of any one of its lateral edges. 15.3 PYRAMIDS Figure 15-8 on page 673 shows some pictures of pyramids. Com- pare Figure 15-8 with Figure 15-1, In what respect does a pyramid differ from a prism? How arc they similar? Since a pyramid is similar in many respects to a prism , some terms for parts or' a prism are also used for parts of a pyramid. We shall use these terms without giving formal definitions. 15.3 Pyramids 673 Figure I5^i Definition 15.7 (See Figure 15-9,) Let Rka polygonal re- gion in a plane o and V a point not in a* For each point P of R there is a segment FV. The union of all such segments is called a pyramid. The polygonal region R is called the hase and V is called the vertex of the pyramid. The distance VT from V to a is the altitude of the pyramid. tx Figure 15*9 For the pyramid in Figure 15-9, AV is a lateral edge and the tri- angular region ABV is a lateral face, Name four other lateral edges and four other lateral faces of the pyramid shown. How many edges lateral and base) does Ida pyramid ki\v m ,,11- IIow main laces in all? How many vertices in all? A cross section of a pyramid is the intersection of the pyramid with a plane parallel to the l*ase provided the intersection contains more than one point. 674 Areas and Volumes of Solids Cfneiir 15 THEOREM 15.4 The boundary of each cross s cction of a tria agu - tar pyramid is a triangle similar to the boundary of the base, and the areas of any two cross sections arc proportional to Lhe squares of the distances of their planes from the vertex of the pyramid. Proof: Tjet the triangular region ABC in plane a he the base of the pyramid as shown in Figure 15-10. LcL fi be a plane parallel to a and intersecting AV, BV, and EV in distinct points A\ W t and C. respec- tively. Then the triangular region A'B'C is a ctoss section gf the pyra- mid. Let S be the area of A ABC, let S' be the area of AA'B'C, let Jt be the distance from the vertex lo the cross section plane, and let h be the altitude of the prism. In Figure 1510, Jt = VF and k = VP. k = VP' h = VP Figure 15-10 To complete the proof of the theorem we shall prove statements 1 and 2. L AA'B'C ~ AABC 2. (S\S) = (**,**) If/? = «,lhenA'=A.H' = B, C = C, Y = F t and k = h. There- fore AA'B'C - AABC and hence A A'B'C s AABC. It follows that AA'B'C' - AABC and that S' = S, k = h, and (S', S) = (&*, *■). Suppose, then, that £ ^ «. We have V 2 A, F A VF coplanar (Why?) with A-A'-V and F-F-V. AT ± Wand A*F i VF. Why? Therefore 15.3 Pyramid* 675 AA'FV- A APV by the A.A. Similarity Theorem. (lA'VF^lAVP LA'VV= ZAPV/i Hence (VA' t VA) = {k. h). In the same way, we can show that AB'FV — ABFV and hence (VB-, VB> = (*, /i). Therefore £VA' T VA) = (VB' t VB) and AA'W - AAVi* by the S. A, S. Similarity Theorem. Therefore (A'B',AB) = (VA\VA) and I A'B\ AS) = (ft, h) A'B' = 4'AJ3. In the same way it can be shown that ft h = £*8C and CA' = Then C'A' = ^>CA. {A'B f , B'C, CA') = (AB, BC, CA) and it follows that AA'B'C <- AABC by the S. S. S. Similarity Theo- rem. Recall that in Chapter 10 it was proved (Theorem 10.15) that if two triangles arc similar, then their areas are proportional to the square of the lengths of any two corresponding sides. Therefore (S\ S) s ((A'/?)*, (AB)*). Since (A'B 1 , AB) = (ft, h) t it follows that {(A'l?)\ (AB)*) = (ft*, V). Therefore (S\ S) = (ft». M)» and the proof is complete. 676 ATM* and Volumes of Solids Chapter 15 Compare our next theorem for pyramids with Theorem 15.2 for prisms. THEOREM 15.5 In any pyramid the areas of any two cross sec- lions arc proportional to the squares of the distances of their planes from die vertex of the pyramid. Proof: Let a pyramid be given as shown in Figure 15-1 1, (We have shown a pyramid whose base is a polygonal region consisting of five sides. The following argument can be modified to apply to a pyramid whose base is a polygonal region consisting of n sides, where n > 3.) k m VP' h=VP Figure 15-11 B Let the polygonal region ABCDE be the base and let A'WC&W be any cross section of the pyramid* Let S be the area of Ihe region ABCDE and let S' be the area of the region A'B'CD'E'. Let k be the distance from the vertex to the cross section plane and let h be die alti- tude of the pyramid. Hie region ABCDE can be divided into nonover- lapping triangular regions tu t 2 , h as shown in Figure 15-1 L Let t{, t' 2 , t$ be die corresponding triangular regions mA'E'CD'F/. Let Tt. 7 a, T s be the areas of t u r 2 , # y , respectively, and let If, Ti, 'ft be the areas of t{, <2» fii, respectively. Thci s- = n + n + r 3 S = T, + T 2 + T; 3- 15.3 Pyramids 677 By Theorem 15.4, From the product property of a proportion it follows that Tift* = TiR Similarly, and T 2 h* = T 2 fc* r s h* = raft?. Then adding and using the Distributive Property, we get pi + n + n)h £ = (Tt + r 3 + rajtf S'Ji* = Sfc* (S\ S) = (fc 2 , **) and this completes the proof. We now use Theorem 15.5 to prove our next theorem. THEOREM 15.6 (The Pyramid Crow Section Theorem) If two pyramids have equal altitudes and if their bases have equal areas, then cross sections equidistant from the vertices have equal areas. Proof: Let two pyramids be given as shown in Figure 15-12. (We have shown pyramids whose bases are triangular regions. The theorem, how- ever, is not restricted to this case and our proof applies equally as well to pyramids whose bases are polygonal regions with more than 3 sides.) Figure 1!S- 12 Areas and Volumes of Solids Chapter 15 Let the base area of each pyramid be S, let h be the altitude of each, and let k be the distance from the vertex to the plane of the cross section of each. Let the areas of the cross sections be Si and S2. We must prove that Sj = Sg. By Theorem 15.5, Therefore (Si. S) f (*t, A*) = (fig, S). Si = s 2 , and the proof is complete. EXERCISES 15.3 In Exercises 1-13, copy and complete each statement. Exercises 3-13 refer Lo Ore pyramid shown in Figure 15-13 in which P is a point in a, the phi n ltd the base, such that f la, Figure 15- IS 1. All the lateral faces of & pyramid are [7] regions. 2. If the boundary of die base of a pyramid is an equilateral triangle, then the boundary of each cross section is |Tj. 3. The region ABCD is called the ]T] of the pyramid. 4. V is called the G] of the pyramid. 5- A V is called a [?) of the pyramid* 6, There are [?] lateral edges in all 7. Counting the lateral edges and the edges of the bases, there are \T\ edges in all. 8. The triangular region AAV is called a of the pyramid. 9, There are [fj lateral faces in all. 15.3 Pyramids 679 10- Counting the base, the total number of faces is [TJ. 11. There are \J} vertices in all. 12. If Vis the number of vertices, K the number of edges, and F the number of faces, then V - E + P = |7J. 13- VP is the [T] of the pyramid, 14. Compare your answer to Exercise 12 with that of Exercise 14 in Exer- cises 15.2. Are they the same? 15. Compute V - E + F(scc Exercise 12) for the pyramid shown in Figure 15-11 and for the prism shown in Figure 15-5, Are your answers the same for both solids? Do you think V — E + F = 2 for every prism or pyramid? Try some more examples using figures from this chapter. 16. Recall that the center of a regular polygon is the center of the circum- scribed circle. The center of a regular polygonal region is the center of the polygon hounding the region. A pyramid is a regular pyramid if and only if its base is a regular polygonal region and the foot of the perpen- dicular from its vertex to its base is the center of the base. Figure 15-13 shows a regular pyramid whose base is a square. Prove that the bound- ary of the lateral face A VB is an isosceles triangle. (Hint: Draw PA and PBand prove that &AFV m ABPVby the S. A. S. Postulate.) 17. Given that the pyramid shown in Figure 15-13 is a regular pyramid whose base is a square region (see Exercise 16), prove that AAVB m ABVC. 18. Draw a picture of a regular pentagonal pyramid, that is, a regular pyra- mid whose base is a regular pentagonal region (see Exercise 16). Label the vertex Vand the vertices of the base A, B t C, D t E. Pick any two lateral faces and prove that they arc congruent isosceles triangles. (Hint: Draw the perpendicular from V to the center of the base as in Figure 15-33.) 19. Recall that corresponding altitudes of congruent triangles are con- gruent. Prove that the lateral surface area of the regular pyramid shown in Figure 15-13 is given by S = £ap, where p is the perimeter of the base and a is the length of the segment whose ondpoints are the vertex of the pyramid and die fool of the perpendicular from the vertex to an edge of the base, (See Exercises 16, 17, and 18.) 20. Let two pyramids, one triangular and one hexagonal, with equal base areas be given. The altitude of each pyramid is 9 in. The cross section of the triangular pyramid that is 3 in. from the base has an area of 40 sq. in. What is the area or the cross section of the hexagonal pyramid that is 3 in. from its lrase? 21. The area of the base of a pentagonal pyramid is 1024. The distance from the vertex to the plane of a cross section is 3 and the altitude of the pyra- mid is 8. Find the area of the cross section. ; J Jink Use Theorem 15.5.) 680 Art** and Volumes of Solids Chapter 15 22, The boundary of the base of a pyramid is an equilateral triangle, and the boundary of each lateral face is an equilateral triangle with sides of length 8. Find the total surface area of the pyramid. 23, The boundary of the base of a pyramid is a square whose sides are 10 em. in length, and the boundary of each lateral face is an equilateral triangle. Find the total surface area of the pyramid. 24, Find the altitude of the pyramid of Exercise 23, 25, Find the altitude of the pyramid of Exercise 22. 26, The area of a cross section of a pyramid is 125. The area of the base is 405. The altitude of the pyramid is 9. Find the distance k from the vertex to the plane of the cross section. fr. Given (a, b) = (c, d) t prove that (a*, fe 2 ) = (c 2 , <#). If (<j, b) = (c, d) with constant of proportionality t> then what is the constant of pro- portionality for {«*, £*) = ( C 2, tP)? Refer tr> the proof of Theorem 15.4. In which sentence of this proof is this property of a proportionality used? 15.4 AREAS AND VOLUMES OF PRISMS AND CYLINDERS As stated in the inrrixhn.-i inn i<> this chapter, we shall not t-cat areas and volumes of solids as rigorously as we did areas of polygonal regions. The concepts of surface area and volume are natural extensions of the area concept developed in Chapter 9. We accept without proof the fact that each solid discussed in this chapter has a surface area and a volume. For volumes, we accept two postulates and use them to prove the volume formulas for prisms, cylinders, pyramids, and cones. Later in the chapter we develop a formula for the volume of a sphere. For surface areas of cylinders, cones, and spheres, we develop the formulas informally and then state them formally as theorems whose proofs are beyond the scope of this book. Our first postulate is similar to Postulate 28, Lhe Rectangle Area Postulate of Chapter 9. POSTULATE 31 (Rectangular Parallelepiped Volume Postu- late) The volume of a rectangular parallelepiped is the product of the altitude and the area of the base. 15.4 Areas and Volumt* of Prisms and Cylinders 681 Figure 15-14 is a picture of a rectangular parallelepiped. By defini- tion, its base is a rectangular region and each of its faces is a rectangular region. Thus any one of its faces could be called the base, and the length of any one of its edges that is perpendicular to that face could be called the altitude. Hfur* 1544 By Postulate 31, the volume Vof the rectangular parallelepiped shown in Figure 15-14 is given by the formula V=Sh, where S is the area of the base and h is the altitude. By Postulate 28, S = oh. Therefore we have V = abh as a formula for the volume of a rectangular parallelepiped. Suppose that an ordinary deck of playing cards is arranged so that its lateral faces are vertical as suggested in Figure 15-15a. It seems rea- sonable dial if the shape of the deck is changed as in Figure 15-15b, the volume of the deck remains the same. It also seems reasonable that the fifteenth card from the bottom in Figure 15- 15a has the same volume as the corresponding card in Figure 15-15b. Figure IS- IS 682 Areas and Volumes of Solids Chapter 15 As a second example,, suppose that we make an approximate model of a rectangular pyramid by forming a slack of thin cards as suggested in Figure 15-16. Of course, the thinner we make the cards, the more cards there will be and the closer the approximation will be. Suppose that the cards in die model are kept at the same level but are allowed to change position by sliding along each other. Figure 15-16 Then the shape of the model changes, but its volume does not change. For different positions of the cards, the base area and the thickness of any two cards at the same level are the same, hence their volume is the same. The total volume of the model is the total volume of the cards, and the total volume does not change when the cards slide along each other. More generally, imagine two solids with equal altitudes and with bases in the same horizontal plane. Suppose that every two cross sec- tions of these solids at equal distances from the bases have equal areas. Then it seems reasonable that the two solids should have equal vol- umes. The reason is that if we imagine die solids being eut into thin slices by planes parallel to the bases, the volumes of slices at the same distance from the bases will be approximately equal. Therefore the volumes of the two solids should be equal. The principle that we have tried to make plausible here is called Gavalieri's Principle after Professor Bonaventum CavaKeri (1598- 1647) of the University of Bologna. He used this principle m obtaining some results that we now find in the calculus. We state this principle, formally as the second postulate of this section. POSTULATE 32 (Cflw/iWi Principle) If two solids have equal altitudes, and if cross sections of these solids at equal distances from the bases have equal areas, then the solids have equal volumes. 15.4 Area* and Volumes of Prisms and Cylinders 683 Cavalicri's Principle is the key to calculating volumes other than rectangular parallelepipeds. We use the principle in the proof of our next theorem, THEOREM 15. 7 The volume V of any prism is the product of it? altitude h and the area S of its base, that is, V = Sfc. Let a prism with altitude h and base area S be given as shown in Figure 15-17a, Let the rectangular parallelepiped shown in Figure 15-t7b have the same altitude h and the same base area S T and let its base and the base of the given prism be in the same plane. Figure 15*17 By the Prism Cross Section Theorem, all cross sections for both prisms have the same area S. By CavalierTs Principle, the two prisms have the same volume. It follows from the Rectangular Parallelepiped Volume Postulate that V = Sh for the given prism. Example I The base boundary of a prism is an equilateral triangle 8 in. on a side. Its altitude is 12 in. Find the volume of the prism. Solution: The volume of the prism is given by V = $h t where S is the area of the triangular base and h is the altitude. We have S = 16 \/3 (show this) and h = 12. Therefore V = 16- V3*I2 = 192 V3, and the volume is 192 v3"cu. in. 684 Areas and Volumes of Solids Ch»pt«r 15 Figure 15-18 shows a diagram of a circular cylinder. Many of the terms used in defining a prism apply equally as well to a circular cylin- der. The base of a prism is a polygonal region. Figure 15-1* The base of a circular cylinder is a circular region, that is, the union of a circle and its interior. Make appropriate changes in the wording of the definition of a prism (Definition 15.1), refer to Figure 15-18, and define a circular cylinder. There are other cylinders in addition to circular cylinders, that is, cylinders whose bases are not circular regions; but we shall consider only circular cylinders in this text. Therefore, when we speak of a cylinder, we mean a circular cylinder. If QQ' in Figure 15-18 is perpendicular to a, then the cylinder is called a right circular cylinder, The altitude, bases, and cross sections of a cylinder are defined in (he same way as are the corresponding parts of prisms. The following two theorems are analogous to Theorems 15.1 and 15,2 for prisms and can be proved in a similar way. We omit the details of the proofs. T1IEOR FM 15.S Hi© bou udary of each cross section of a cyl i n der is a circle that is congruent to the boundary of the base. Outline of Proof: (See Figure 15-19 on page 685.) Let R be the lyase of the given cylinder and let H' be any cross section of the cylinder- Let C be the center of the circle that bounds the base, let P be a point on that circle, and let C and F be the corresponding points in R' t Let 15.4 Areas «nd Volumes of Prisms and Cylinders 685 CP = t and let CF = t>, Then PCCF is a parallelogram (Why?), and FC = r f = PC = r. Since PC has a constant value regardless of the position of P on the base circle, then FC has a constant value. Thus all points F lie on a circle with radius r* and center U. Figure 15-19 Therefore the cross section is a circular region and its boundary is a circle of radius r. Then its boundary is congruent to the boundary of the base. THEOREM 15.9 (The Cylinder Cross Section Theorem) The area of a cross section of a cylinder is equal to the area of the base. Proof: Assigned as an exercise* Cavalieri's Principle is used in the proof of the following theorem on the volume of a cylinder. THEOREM 1 5 JO The volume of a cylinder is the product of the altitude and the area of the Imss. Proof: The proof is similar to that of Theorem 15.7 and is assigned as an exercise. Imagine slitting a right circular cylinder and unrolling its lateral surface onto a plane. Which figure is obtained? Figure 15-20 on page 081 BBggesis that the boundary of the region khufl obtained k a sec- tangle whoso altitude is the altitude of the cylinder and whose base is the circumference of the base of the cylinder. Thus, if -r is the radius sac Areas, and Volumes of Solids Chapter 15 of the boundary of the base of a right cylinder and h is the altitude, then the lateral surface area of the cylinder is equal to the area of the rectangular region obtained by "unrolling" the cylinder, that is, the lateral surface area is 2vrh, Since the area of each of the circular bases is tff 2 , the total surface arm of a right cylinder is 27rrh + 2irA -L Figure 1 5-3)0 We state these results formally as our next theorem. THEOREM 15.11 The lateral surface area of a circular cylinder of base radius r and altitude h is 2irrh and its total surface area is 2irrk + Swr 6 . liliniiflb 2 Find the total surface area of a cylinder if the radius of the circle that bounds the base is 7 and the altitude of the cylinder is 10. Solution; The lateral surface area is 2irrh= 2ff-7« 10 = Mto The area of each base is at* = ??(7) 2 = 49tt. Therefore the total surface area is 140*7 + 2-49* = 140^ + 9&r = 23&r. EXERCISES 15.4 1, A rectangular tank 6 ft, by 4 ft. is used for watering horses. If the tank is filled with water to a depth of 3 ft., how many cubic feet of water are in the tank? 2* One gallon of water occupies 231 cu. in, of space- To the nearest hun- dredth, how many gallons of walcr are contained in a space of 1 cu. ft.'? 3. To the nearest gallon, how inany gallons of water are in the tank of Exercise 1? (Sec Exercise 2.) 4, Show that the volume V of a cube of side e is given by V = *3. 15,4 Ar»a* and VoJumw of Prhmt and Cylinders 687 5. Find the volume of a cube whose edge is 6 in, Find its total surface area. 6. Write a formula for the volume Vof a cylinder if its altitude is h and the radius of tts base circle is r. 7. A water tank in the shape of a cylinder is 40 ft. high. The diameter of its base is 2S ft. Find the volume of the tank. 8. To the nearest thousand gallons, how many gallons of water will the tank of Exercise 7 hold? (Use w = -^- and see Exercise 2.) 9. The altitude of a cylinder is 8 in. and the diameter of its ba.se circle is 3 in. Kind the volume and total surface area of the cylinder. 10. On a shelf in Roy's supermarket there are two cylindrical cans of coffee. The first is 1 j times as tall as the second, but the second has a diameter 1^ times that of the first. How should Roy price the second can in rela- tion to the first if he wants the price per unit of volume to be the same for both cans? 11. How do the volumes of two cylinders compare if their altitudes are the same but the radius of the base circle of the second cylinder is three times that of the first? 12. How do die volumes of two cylinders compare if the radii of their bass circles are the same but the altitude of the second cylinder is throe times that of the first? 13. Draw a suitable figure and prove Theorem 15.9. 14 I3raw a suitable figure and prove Theorem 15.10, 15. A brick chimney In the form of a cylindrical shell and 2o ft. tall is to be built. The inside and outside diameters are 24 in. and 10 in., respec- tively. If it takes 31 bricks per cubic foot of chimney, find die approx- imate number of bricks needed (Use v = 3.14.) 16. An air conditioning unit is to be installed in a rectangular building. In order to install the correct size unit, it is necessary to know the number of cubic feet of air inside the building. If the dimensions of the building are as shown in Figure 15-21 , find the volume of air inside the building. 17. A block of wood in the shape of a cube has edges 16 in. in length. A circular hole 7 in. in diameter is bored through the block from top to bottom. Find die volume of the part of the block that remains. 18. In Exercise 17, if e is the length of die edge of the cube and r is the radius of the circular hole, write ft formula for the volume Vof die block that remains after the hole has been bored. HI Areas and Volumes of Solids Chapter 15 15.5 VOLUMES OF PYRAMIDS AND CONES Here wc develop formulas for calculating volumes of pyramids and cones and for the surface area of a cone. As in Section 15,4 , Cavalieri's Principle plays a key role in the proofs of theorems on volume. THEOREM 15,12 Two pyramids with the same altitude and the same base area have the same volume. Proof: Let two pyramids lws given as suggested in Figure 15-22. By the Pyramid Cross Section Theorem, corresponding cros